Note that this sum is in fact
$$2\sum_{m\ge 1} \frac{\coth m\pi}{m^{4p+3}}.$$
The sum term $$T_p(x) =
2\sum_{m\ge 1} \frac{\coth mx}{m^{4p+3}}$$
is harmonic and may be evaluated by inverting its Mellin transform.
Observe that $$\coth x = \frac{e^x+e^{-x}}{e^x-e^{-x}}
= 1 + 2\frac{e^{-x}}{e^x-e^{-x}}
= 1 + 2\frac{e^{-2x}}{1-e^{-2x}}.$$
This yields
$$T_p(x) = 2\zeta(4p+3) +
4\sum_{m\ge 1} \frac{1}{m^{4p+3}} \frac{e^{-2mx}}{1-e^{-2mx}}.$$
We will now work with
$$S(x) = \sum_{m\ge 1} \frac{1}{m^{4p+3}}
\frac{e^{-2mx}}{1-e^{-2mx}}$$
and establish a functional equation for $S(x)$ that has $x=\pi$ as a
fixed point.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^{4p+3}},
\quad \mu_k = k \quad \text{and} \quad
g(x) = \frac{\exp(-2x)}{1-\exp(-2x)}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$
which we compute as follows:
$$\int_0^\infty \frac{\exp(-2x)}{1-\exp(-2x)} x^{s-1} dx
= \int_0^\infty \sum_{q\ge 1} \exp(-2qx)\; x^{s-1} dx
\\ = \sum_{q\ge 1} \int_0^\infty \exp(-2qx)\; x^{s-1} dx
= \Gamma(s) \sum_{q\ge 1} \frac{1}{2^s q^s}
= \frac{1}{2^s} \Gamma(s) \zeta(s).$$
with fundamental strip $\langle 1,\infty \rangle.$
Hence the Mellin transform $Q(s)$ of $S(x)$ is given by
$$ Q(s) = \frac{1}{2^s} \Gamma(s) \zeta(s) \zeta(s+4p+3)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s+4p+3)$$
where $\Re(s+4p+3) > 1$ or $\Re(s) > -4p-2$.
Intersecting the fundamental strip and the half-plane from the zeta
function term we find that the Mellin inversion integral for an
expansion about zero is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate in the left half-plane $\Re(s)<3/2.$
The plain zeta function term cancels the poles of the gamma function
term at even negative integers $2q\le -2$ and the compound zeta
function term the poles at odd negative integers $2q+1\le -4p-5$. We
are left with the contributions from $s=1$, $s=0$ and $s=-1$ which are
$$\begin{align}
\mathrm{Res}(Q(s)/x^s; s=1) & = \frac{1}{2x}\zeta(4p+4) \\
\mathrm{Res}(Q(s)/x^s; s=0) & = -\frac{1}{2}\zeta(4p+3) \\
\mathrm{Res}(Q(s)/x^s; s=-1) & = \frac{1}{6} x\zeta(4p+2).
\end{align}$$
The remaining contributions are
$$\sum_{q=1}^{2p+1} \mathrm{Res}(Q(s)/x^s; s=-2q-1)
\\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1}
\frac{(-1)^{2q+1}}{(2q+1)!}
\zeta(-2q-1) \zeta(4p+2-2q)
\\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1}
\frac{1}{(2q+1)!}
\frac{B_{2q+2}}{2q+2}
(-1)^{2p+1-q+1} \frac{B_{4p+2-2q}(2\pi)^{4p+2-2q}}{2(4p+2-2q)!}
\\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1}
\frac{B_{2q+2}}{(2q+2)!}
(-1)^{-q} \frac{B_{4p+2-2q}(2\pi)^{4p+2-2q}}{2(4p+2-2q)!}
\\ = \sum_{q=2}^{2p+2} 2^{2q-1} x^{2q-1}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}(2\pi)^{4p+4-2q}}{2(4p+4-2q)!}
\\ = 2^{4p+2} \sum_{q=2}^{2p+2} x^{2q-1}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}\pi^{4p+4-2q}}{(4p+4-2q)!}
\\ = 2^{4p+2} \sum_{q=2}^{2p+2} x^{2q-1} \pi^{4p+4-2q}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}$$
and the one from the pole of the compound zeta function term at
$s=-4p-2$ which we'll do in a moment.
Now some algebra shows that setting $q=0$ and $q=1$ in the sum
produces precisely the values that we obtained earlier for the poles
at $s=1$ and $s=-1$ so we may extend the sum to start at zero, keeping
only the residue from the pole at $s=0$ to get
$$2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$
We have thus established that
$$S(x) = -\frac{1}{2}\zeta(4p+3)
+ \mathrm{Res}(Q(s)/x^s; s=-4p-2)
\\ + 2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}
\\ + \frac{1}{2\pi i}
\int_{-4p-4-i\infty}^{-4p-4+i\infty} Q(s)/x^s ds.$$
To treat the integral recall the duplication formula of the gamma
function:
$$\Gamma(s) =
\frac{1}{\sqrt\pi} 2^{s-1}
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right).$$
which yields for the integral
$$\int_{-4p-4-i\infty}^{-4p-4+i\infty}
\frac{1}{2\sqrt\pi}
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right)
\zeta(s)
\zeta(s+4p+3)/x^s ds.$$
Furthermore observe the following variant of the functional equation
of the Riemann zeta function:
$$\Gamma\left(\frac{s}{2}\right)\zeta(s)
= \pi^{s-1/2} \Gamma\left(\frac{1-s}{2}\right)
\zeta(1-s)$$
which gives for the integral
$$\int_{-4p-4-i\infty}^{-4p-4+i\infty}
\frac{\pi^{s-1}}{2}
\Gamma\left(\frac{s+1}{2}\right)
\Gamma\left(\frac{1-s}{2}\right)
\zeta(1-s)
\zeta(s+4p+3)/x^s ds
\\ = \int_{-4p-4-i\infty}^{-4p-4+i\infty}
\frac{\pi^{s}}{2}
\frac{1}{\sin(\pi/2(s+1))}
\zeta(1-s)
\zeta(s+4p+3)/x^s ds.$$
Now put $s=-4p-2-u$ to get
$$\int_{2-i\infty}^{2+i\infty}
\frac{\pi^{-4p-2-u}}{2}
\frac{1}{\sin(\pi/2(-4p-2-u+1))}
\\ \times \zeta(u+4p+3)
\zeta(1-u)/x^{-4p-2-u} du
\\ = \frac{x^{4p+2}}{\pi^{4p+2}}
\int_{2-i\infty}^{2+i\infty}
\frac{\pi^{-u}}{2}
\frac{1}{\sin(\pi/2(-(u+1)))}
\\ \times \zeta(u+4p+3)
\zeta(1-u)/x^{-u} du
\\ = - \frac{x^{4p+2}}{\pi^{4p+2}}
\int_{2-i\infty}^{2+i\infty}
\frac{\pi^{-u}}{2}
\frac{1}{\sin(\pi/2(u+1))}
\\ \times \zeta(u+4p+3)
\zeta(1-u)/x^{-u} du.$$
We have established the functional equation
$$S(x) = -\frac{1}{2}\zeta(4p+3)
+ \mathrm{Res}(Q(s)/x^s; s=-4p-2)
\\ + 2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}
-\left(\frac{x}{\pi}\right)^{4p+2} S\left(\frac{\pi^2}{x}\right).$$
Setting $x=\pi$ we have
$$S(\pi) = -\frac{1}{2}\zeta(4p+3)
+ \mathrm{Res}(Q(s)/\pi^s; s=-4p-2)
\\ + 2^{4p+2} \pi^{4p+3} \sum_{q=0}^{2p+2}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}
-\left(\frac{\pi}{\pi}\right)^{4p+2} S\left(\pi\right)$$
and hence
$$S(\pi) = -\frac{1}{4}\zeta(4p+3)
+ \frac{1}{2} \mathrm{Res}(Q(s)/\pi^s; s=-4p-2)
\\ + 2^{4p+1} \pi^{4p+3} \sum_{q=0}^{2p+2}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$
To conclude we treat the residue that we have defered until now.
Recall the alternate form of $Q(s)/\pi^s,$
$$\frac{\pi^{s}}{2}
\frac{1}{\sin(\pi/2(s+1))}
\zeta(1-s)
\zeta(s+4p+3)/\pi^s
\\ = \frac{1}{2}
\frac{1}{\sin(\pi/2(s+1))}
\zeta(1-s)
\zeta(s+4p+3).$$
It follows that the residue at $s=-4p-2$ is
$$\frac{1}{2}
\frac{1}{\sin(\pi/2(-4p-1))}
\zeta(4p+3)
= \frac{1}{2}
\frac{1}{\sin(-\pi/2)}
\zeta(4p+3) = -\frac{1}{2} \zeta(4p+3).$$
This finally yields for $S(\pi)$
$$S(\pi) =
-\frac{1}{2}\zeta(4p+3)
+ 2^{4p+1} \pi^{4p+3} \sum_{q=0}^{2p+2}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$
Computing $T_p(\pi)$ we thus obtain
$$2\zeta(4p+3)
-4\times\frac{1}{2}\zeta(4p+3)
+ 2^{4p+3} \pi^{4p+3} \sum_{q=0}^{2p+2}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}$$
or
$$(2\pi)^{4p+3} \sum_{q=0}^{2p+2}
\frac{B_{2q}}{(2q)!}
(-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!},$$
QED.
The inspiration for this calculation is from the paper "Mellin
Transform and its Applications" by Szpankowski.
