My solution to this infinite sum involves complex analysis:
Set $f(x)=\frac{1}{(2x+1)^{3}}$ then $\psi(x)=\frac{\pi}{\sin(\pi x)f(x)}$ notice how the reason we chose $\psi (x)$ to be such is that it has poles at every integer such that
$\operatorname{Res}[\psi(x),n]=(-1)^{n}f(n)$ where $n \in \mathbb Z$.
Consider this contour, a box of side length $2(N+1)$:
There is a pole at every integer value and at $z=\frac {-1}{2}$
It is easy to show that in $\lim_{n \to \infty}$, $$\oint_C \psi(z)=0$$
$\Rightarrow$ $\oint_C$ $\psi$(z)dz=$\sum_{i=-\infty}^{\infty}$Res[$\psi$(z),i] + Res[$\psi(z)$,z=$\frac{-1}{2}$]=0
$$\Rightarrow \sum_{n=-\infty}^{\infty}(-1)^{n}f(n)+\frac{1}{2!} \lim_{x \to \frac {-1}2}\frac{d^{2}}{dx^{2}}\frac{\left(x+\frac 12 \right)^{3}.\pi \csc(\pi x)}{(2x+1)^{3}}=0$$
the limit is easy to evaluate so I won't show that
$$\Rightarrow \sum_{n=-\infty}^{\infty}(-1)^{n}f(n)-\frac {\pi^{3}}{2^{4}}=0 $$ since our sum is just twice the sum from n=0 to n=$\infty$, we get $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{3}}=\frac 12.\frac {\pi^{3}}{2^{4}}=\frac{\pi^{3}}{32}$$ I was curious if there is another method to evaluate this infinite sum
