Your formula is a variant of a formula from Euler ($1772$) (exposed in Ayoub's paper Euler and the Zeta Function (1974)
p. $1085$) rediscovered by Ramaswami ($1934$) and more recently by Ewell ($1990$).
See too Adamchik's paper and Boros and Moll's nice book 'Irresisitible Integrals' $(11.4.6)$ :
\begin{align}
\zeta(3)&=\frac{\pi^2}7\left(1-4\sum_{n=1}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\right)\qquad\text{or since}\;\zeta(0)=-\frac12\\
\tag{1}\zeta(3)&=-\frac{4\pi^2}7\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\\
\end{align}
This becomes indeed, after substitution of $\,\displaystyle \frac{\zeta(2n)}{2^{2n}}=(-1)^{n+1}\frac{B_{2n}\,\pi^{2n}}{2(2n)!}$ :
\begin{align}
\zeta(3)&=\frac{2\,\pi^2}7\sum_{n=1}^\infty (-1)^n\frac{B_{2n}\,\pi^{2n}}{(2n+2)(2n+1)(2n)!}\\
\tag{2}\zeta(3)&=\frac27\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\pi^{2n+2}}{(2n+2)!}\\
\end{align}
Let's reproduce here Boros and Moll's proof of $(1)$ and start with the generating function for central binomial coefficients :
$$\tag{3}\frac 1{\sqrt{1-4t}}=\sum_{n=0}^\infty \binom{2n}{n}t^n$$
Setting $\,t:=\left(\dfrac x2\right)^2$ and integrating we get :
$$\tag{4}\arcsin(x)=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{x^{2n+1}}{2n+1}$$
(btw Euler found too this neat expression for the square :
$\displaystyle \arcsin(x)^2=\sum_{n=1}^\infty \frac{2^{2n}}{\binom{2n}{n}}\frac{x^{2n}}{2\,n^2}\;$)
Dividing $(4)$ by $x$ and integrating we get :
$$\tag{5}\int\frac{\arcsin(x)}xdx=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{x^{2n+1}}{(2n+1)^2}$$
For $x:=\sin(t)$ this becomes :
$$\tag{6}\int\frac{t\cos(t)}{\sin(t)}dt=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(\sin(t))^{2n+1}}{(2n+1)^2}$$
At this point we may use the well known generating function for $\zeta(2n)$ :
$$\pi\;x\;\cot(\pi\;x)=1-2\sum_{n=1}^\infty \zeta(2n)\;x^{2n}$$
So that setting $t:=\pi x\,$ and integrating we get using $(6)$ :
$$\tag{7}\int t\,\cot(t)\,dt=t-2\sum_{n=1}^\infty \frac{\zeta(2n)\;t^{2n+1}}{\pi^{2n}(2n+1)}=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(\sin(t))^{2n+1}}{(2n+1)^2}$$
Integrating this again from $0$ to $\frac {\pi}2$ returns :
$$\frac {(\pi/2)^2}2-2\sum_{n=1}^\infty \frac{\zeta(2n)\;(\pi/2)^{2n+2}}{\pi^{2n}(2n+2)(2n+1)}=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{1}{(2n+1)^2}\int_0^{\pi/2}\sin^{2n+1}(t)\,dt$$
The integral at the right may be evaluated with the Wallis formula :
$$\frac{\binom{2n}{n}}{2^{2n}}\int_0^{\pi/2}\sin^{2n+1}(t)\,dt=\frac1{2n+1}$$
and we get
$$\tag{8}\frac {\pi^2}8-\frac{\pi^2}2\sum_{n=1}^\infty \frac{\zeta(2n)}{2^{2n}(2n+2)(2n+1)}=\sum_{n=0}^\infty\frac{1}{(2n+1)^3}=\frac 78\zeta(3)$$
(separating the even and odd terms of zeta it is easy to prove the more general identity $\;\displaystyle\sum_{n=0}^\infty\frac{1}{(2n+1)^m}=\left(1-\frac 1{2^m}\right)\zeta(m)\;$)
From $(8)$ we deduce $(1)$ and thus your formula $(2)$.
GENERALIZATION
Let's define
$$f(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\pi^{2n+m-1}}{(2n+m-1)!}=-\pi^{m-1}\,\sum_{n=0}^{\infty}\frac{(2n)!\;\zeta(2n)}{(2n+m-1)!\,2^{2n-1}}$$
then using the method nicely exposed in your answer (computing further integrals of $\dfrac{x}{e^x-1}$ that is evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-\pi\,i$)
we may obtain :
\begin{align}
f(1)&=0\\
f(2)&=\pi\,\log(2)\\
f(3)&=\frac 72\zeta(3)\\
f(4)&=\frac 54\pi\zeta(3)\\
\tag{9}f(5)&=-\frac {31}4\zeta(5)+\pi^2\zeta(3)\\
f(6)&= \frac{\pi^3}3\zeta(3) -\frac{49}{16}\pi\zeta(5)\\
f(7)&=\frac{381}{32}\zeta(7)-2\pi^2\zeta(5)+\frac{\pi^4}{12}\zeta(3)\\
f(8)&= \frac{\pi^5}{60}\zeta(3)-\frac 23\pi^3\zeta(5)+\frac{321}{64}\pi\zeta(7)\\
f(9)&=-\frac{511}{32}\zeta(9)+3\pi^2\zeta(7)-\frac{\pi^4}6\zeta(5)+\frac{\pi^6}{360}\zeta(3)\\
\end{align}
and so on and thus conjecture that for $\;n>1$ we have : $\qquad\qquad\qquad\qquad\qquad\qquad\qquad(10)$
\begin{align}
f(2n+1)&=(-1)^{n+1}\left[n\left(4-2^{1-2n}\right)\zeta(2n+1)+2\sum_{k=1}^{n-1} \frac{(-1)^k(n-k)}{(2k)!}\pi^{2k}\ \zeta(2(n-k)+1)\right]\\
f(2n+2)&=(-1)^{n+1}\pi\left[\left(2n-1+2^{-2n}\right)\zeta(2n+1)+2\sum_{k=1}^{n-1} \frac{(-1)^k(n-k)}{(2k+1)!}\pi^{2k}\,\zeta(2(n-k)+1)\right]\\
\end{align}
For a proof see the Theorem A of Cvijovic and Klinowski ($1997$) "New rapidly convergent series representations for $\zeta(2n+1)$" with the result (similar to the first equation $(10)$) :
$$\tag{11}\zeta(2n+1)=\frac{(-1)^n\,(2\pi)^{2n}}{n(2^{2n+1}-1)}\left[\sum_{k=1}^{n-1}\frac{(-1)^{k-1}\,k\,\zeta(2k+1)}{(2n-2k)!\,\pi^{2k}}+\sum_{k=0}^\infty\frac{(2k)!\,\zeta(2k)}{(2n+2k)!\,2^{2k}}\right]$$
This allows to find whole families of formulae for $\zeta(2n+1)$ with some neat instances like these :
(cf the comments for $\zeta(5)$)
\begin{align}
\zeta(5)&=\frac{16}{23}\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{(2n+2)\ \pi^{2n+4}}{(2n+5)!}\\
\zeta(5)&=-\frac{2^5\,\pi^4}{23}\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+5)(2n+4)(2n+3)(2n+1)\,2^{2n}}\\
\zeta(7)&=-\frac{2^7\,\pi^6}{4719}\sum_{n=0}^\infty \frac{(128\,n+211)\quad\zeta(2\,n)}{(2n+7)(2n+6)(2n+5)(2n+4)(2n+3)(2n+1)\,2^{2n}}\\
\zeta(9)&=-\frac{2^{10}\,\pi^8}{77259}\sum_{n=0}^\infty \frac{(449\,n+780)\quad\quad\zeta(2\,n)}{(2n+9)(2n+8)(2n+7)(2n+6)(2n+5)(2n+3)(2n+1)\,2^{2n}}\\
\zeta(11)&=-\frac{2^{12}\,\pi^{10}}{\small{395681475}}\sum_{n=0}^\infty\frac{(2497024\,n^2+10676923\,n+11093808)\quad\zeta(2n)}{(2n+11)(2n+10)\cdots(2n+6)(2n+5)(2n+3)(2n+1)\,2^{2n}}\\
\zeta(13)&=-\frac{2^{13}\,\pi^{12}}{\small{75159854595}}\sum_{n=0}^\infty\frac{\small{(619308920\,n^2+2659184244\,n+2771839831)\quad}\zeta(2n)}{\small{(2n+13)(2n+12)\cdots(2n+8)(2n+7)(2n+5)(2n+3)(2n+1)\,2^{2n}}}\\
\zeta(15)&=-\frac{2^{15}\,\pi^{14}}{\small{1697182926260535}}\sum_{n=0}^\infty\frac{\small{(21253850808320\,n^3+165886464354888\,n^2+415352534250460\,n+332739769444737)}\zeta(2n)}{\small{(2n+15)(2n+14)\cdots(2n+8)(2n+7)(2n+5)(2n+3)(2n+1)\,2^{2n}}}\\
\end{align}
The coefficients for these formulae may possibly be obtained using recurrences (by rewriting $(11)$ or using alternative relations) but I obtained them numerically with high precision (i.e. these identities are only conjectured).
We may compare them with the Euler formula $(1)$ and its variants :
\begin{align}
\zeta(3)&=-\frac{4\pi^2}7\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+2)(2n+1)\,2^{2n}}\\
\zeta(3)&=-\frac{8\pi^2}5\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(2n+3)(2n+2)(2n+1)\,2^{2n}}\\
\zeta(3)&=\frac{2\pi^2}7\left(\log(2)+\sum_{n=0}^\infty \frac{\zeta(2\,n)}{(n+1)\,2^{2n}}\right)\\
\end{align}
These two last formulae were obtained by Chen and Srivastava in "Some Families of Series Representations for the Riemann $\zeta(3)$" and reproduced in Srivastava and Choi's book "Zeta and q-Zeta Functions and Associated Series and Integrals" (see p.$405$ "$4.2$ Rapidly Convergent Series for $\zeta(2n
+ 1)$") that contains many other series for $\zeta$ and general formulae like this one :
\begin{align}
\zeta(2n+1)&=\frac{(-1)^{n-1}(2\pi)^{2n}}{2^{2n+1}-1}\left[\frac{H_{2n}-\log(\pi)}{(2n)!}+\sum_{k=1}^{n-1}\frac{(-1)^k\,\zeta(2k+1)}{(2n-2k)!\,\pi^{2k}}+2\sum_{k=1}^\infty\frac{(2k-1)!\,\zeta(2k)}{(2n+2k)!\,2^{2k}}\right]\\
\end{align}
A review is proposed in Katsurada's paper "Rapidly convergent series representations for $\zeta(2n+1)$ and their $\chi$-analogue".
OTHER GENERALIZATIONS
The OP asked what would happen with the upper bound of the integral $x=-\pi\,i\,$ replaced by $\,x=-\frac{\pi}2i\,$ so let's investigate this by defining
$$g(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{\left(\frac{\pi}2\right)^{2n+m-1}}{(2n+m-1)!}=-\left(\frac{\pi}2\right)^{m-1}\,\sum_{n=0}^{\infty}\frac{2\,(2n)!\;\zeta(2n)}{(2n+m-1)!\;2^{4n}}$$
then evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-\frac{\pi}2\,i\;$ returns if $\;\displaystyle \beta(s):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet beta function :
\begin{align}
g(1)&=\frac{\pi}4\\
g(2)&=\frac{\pi}4\log(2)+\beta(2)\\
g(3)&=\frac{35}{16}\zeta(3) -\frac 12\pi\,\beta(2)\\
\tag{12}g(4)&=\frac{61}{64}\pi\,\zeta(3)-3\,\beta(4)\\
g(5)&=\frac 14\pi^2\,\zeta(3)-\frac{527}{128}\zeta(5)+\frac 12\pi\,\beta(4)\\
g(6)&=\frac 1{24}\pi^3\,\zeta(3)-\frac{2033}{1024}\pi\,\zeta(5)+5\,\beta(6)\\
g(7)&=\frac 1{192}\pi^4\,\zeta(3)-\frac 12\pi^2\,\zeta(5)+\frac{24765}{4096}\zeta(7)-\frac 12\pi\,\beta(6)\\
\end{align}
and so on so that we may conjecture that for $\;n>1$ : $\qquad\qquad\qquad\qquad\qquad\qquad\qquad(13)$
\begin{align}
g(2n+1)&=(-1)^n\left[\frac {\pi}2\beta(2n)-n\frac{2^{4n+1}+2^{2n}-1}{2^{4n}}\zeta(2n+1)-\\2\sum_{k=1}^{n-1}(-1)^k\frac{n-k}{(2k)!}\left(\frac{\pi}2\right)^{2k}\zeta(2(n-k)+1))\right]\\
g(2n+2)&=(-1)^n\left[(2n+1)\beta(2n+2)-\left(n-\frac{2^{2n}-1}{2^{4n+2}}\right)\pi\,\zeta(2n+1)-\\2\sum_{k=1}^{n-1}(-1)^k\,\frac{n-k}{(2k+1)!}\left(\frac{\pi}2\right)^{2k+1}\zeta(2(n-k)+1)\right]\\
\end{align}
Returning some expressions of interest for the Catalan constant $G$ and other $\beta(2n)$ constants :
\begin{align}
G=\beta(2)&=-\frac{\pi}4\log(2)-\pi\,\sum_{n=0}^{\infty}\frac{\zeta(2n)}{(2n+1)\;2^{4n}}\\
\beta(4)&=-\frac {\pi^3}{840}\left[\frac{61}4\log(2)+\sum_{n=0}^\infty \frac{(244(n+2)(n+1)-9)\,\zeta(2n)}{(2n+3)(2n+2)(2n+1)\,2^{4n}}\right]\\
\beta(6)&=-\frac{\pi^5}{3541440}\left[\frac{50345}8\log(2)+\\\sum_{n=0}^\infty\frac{(402760n^4 + 3020700n^3 + 8300346n^2 + 9777801n + 4077233)\zeta(2n)}{(2n+5)(2n+4)(2n+3)(2n+2)(2n+1)\,2^{4n}}\right]\\
\end{align}
as well as fast series for $\zeta(\text{odd})$ :
\begin{align}
\zeta(3)&=-\frac{2\;\pi^2}{35}\left[\log(2)+4\sum_{n=0}^\infty\frac{(2n+3)\quad\zeta(2n)}{(2n+2)(2n+1)\;2^{4n}}\right]\\
\zeta(5)&=-\frac{2\;\pi^4}{55335}\left[157\,\log(2)+8\sum_{n=0}^\infty\frac{(628\,n^3+3140\,n^2+5111\,n+2581)\;\zeta(2n)}{(2n+4)(2n+3)(2n+2)(2n+1)\;2^{4n}}\right]\\
\end{align}
We could get slower converging results with the upper bound of the integral replaced by $\,x:=-2\pi\,i$. For this let's define
$$h(m):=\sum_{n=0}^{\infty}(-1)^nB_{2n}\frac{(2\,\pi)^{2n+m-1}}{(2n+m-1)!}=-(2\,\pi)^{m-1}\,\sum_{n=0}^{\infty}\frac{2\,(2n)!\;\zeta(2n)}{(2n+m-1)!}$$
then evaluating $\;\displaystyle \int \cdots \int \frac{x}{e^x-1} dx\cdots\,dx=\int_0^x \frac{t}{e^t-1}\frac{(x-t)^{n-1}}{(n-1)!} dt\,$ for $x=-2\,\pi\,i$)
we may obtain :
\begin{align}
h(3)&=0\\
h(4)&=6\,\pi\,\zeta(3)\\
h(5)&=4\,\pi^2\,\zeta(3)\\
\tag{14}h(6)&=\frac 83\pi^3\,\zeta(3)-10\,\pi\,\zeta(5)\\
h(7)&=\frac 43\pi^4\,\zeta(3)-8\,\pi^2\,\zeta(5)\\
h(8)&= \frac 8{15}\pi^5\,\zeta(3)-\frac{16}3\pi^3\,\zeta(5)+14\,\pi\,\zeta(7)\\
h(9)&= \frac 8{45}\pi^6\,\zeta(3)-\frac 83\,\pi^4\,\zeta(5)+12\,\pi^2\,\zeta(7)\\
h(10)&=\frac{16}{315}\pi^7\,\zeta(3)-\frac{16}{15}\pi^5\,\zeta(5)+8\,\pi^3\,\zeta(7)-18\,\pi\,\zeta(9)\\
\end{align}
And find further expressions for $\zeta(\text{odd})$ (using these expressions or combining them with the earlier results) but since the subject appears endless I'll stop here.
