Our goal is to show
$$\sum_{n=0}^{\infty}
\frac{1}{(2n+1)^2\sinh\left ( \frac\pi2(2n+1) \right )^2}
=\frac14S +\frac{\pi^2}{48}-\frac{3\pi}2\ln(2)+2G.$$
The First Step.(as a hint) We consider
$
\sum_{m\ge0,n\ge1}n^3e^{-\pi xn(2m+1)}
=\frac{1}{16}\vartheta_2(q)^4\vartheta_3(q)^4.
$
Therefore it's easy to show
$$
\sum_{n=0}^{\infty}
\frac{1}{(2n+1)^2\sinh\left ( \frac\pi2(2n+1) \right )^2}
=\frac{\pi^2}{4} \int_{1}^{\infty}(x-1)\vartheta_2(q)^4\vartheta_3(q)^4
\text{d}x,
$$
which is solved to be the aforementioned hypergeometric series.
On the one hand, by integrating $f_1(z)=\frac{\pi \tan(\pi z)}{z^2\sinh(\pi z)^2},f_2(z)=\frac{\pi \cot(\pi z)\tanh(\pi z)}{z^3}$ and residue theorem, gives
\begin{aligned}
&\sum_{n=1}^{\infty} \left ( \frac2\pi
\frac{\tanh(\pi n)}{n^3}-\frac{1}{n^2\cosh(\pi n)} \right )
=S+\frac{\pi^2}{12}-6\pi\ln(2)+8G,\\
&\sum_{n=0}^{\infty} \frac{\coth\left ( \frac\pi2(2n+1) \right ) }{(2n+1)^3}
=\frac{\pi^3}{24} -\frac18\sum_{n=1}^{\infty}
\frac{\tanh(\pi n)}{n^3}.
\end{aligned}
On the other hand, compute
$
\sum_{(m,n)\in\mathbb{Z}^2}\frac{1}{\left [
\left ( m+\frac12 \right ) ^2+n^2\right ]^2 } =4\pi^2G
$,
from which we could find
$$
\sum_{n=1}^{\infty} \left ( \frac2\pi
\frac{\tanh(\pi n)}{n^3}-\frac{2}{n^2\cosh(\pi n)} \right )
=8G-\frac{2\pi^2}3.\square
$$
Here are some relative results.
Entry 1.
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^3}
=\frac{7\pi^3}{180},\\
&\sum_{n=1}^{\infty} \frac{1}{n^3\left ( e^{2\pi n}-1 \right )
} =\frac{7\pi^3}{360}-\frac{1}{2}\zeta(3),\\
&\sum_{n=1}^{\infty} \frac{1}{n^2\sinh(\pi n)^2}
=\frac23G-\frac{11\pi^2}{180},\\
&\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n\sinh(\pi n)^2}=\frac\pi{30}-\frac{G}{3\pi}.
\end{aligned}
Entry 2.
\begin{aligned}
&\sum_{n=1}^{\infty} \frac{\tanh(\pi n)}{n^3}
=\pi S+\frac{5\pi^3}{12}-6\pi^2\ln(2)+4\pi G,\\
&\sum_{n=1}^{\infty} \frac{1}{n^3\left ( e^{2\pi n}+1 \right )
} =\frac{1}{2}\zeta(3)-\frac{\pi}{2} S
-\frac{5\pi^3}{24}+3\pi^2\ln(2)-2\pi G,\\
&\sum_{n=1}^{\infty} \frac{1}{n^2\cosh(\pi n)^2}
=S+\frac{3\pi^2}{4}-6\pi\ln(2),\\
&\sum_{n=1}^{\infty} \frac{\tanh(\pi n)}{n\cosh(\pi n)^2}=-\frac\pi2+\frac32\ln(2)+\frac{2G}\pi-\frac1{32}\,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ),
\end{aligned}
where $S=\frac{7}{2}{}_6F_5\left ( \frac34,\frac34,1,1,1,\frac{15}8;
\frac78,\frac54,\frac54,2,2;1 \right )
+\frac{\pi}{8}{}_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right )$.
Entry 3.
\begin{aligned}
&\sum_{n=0}^{\infty} \frac{\tanh\left ( \frac\pi2(2n+1) \right ) }{(2n+1)^3}
=\frac{\pi^3}{32} ,\\
&\sum_{n=0}^{\infty}\frac{1}{(2n+1)^3\left ( e^{\pi(2n+1)}+1 \right ) }
=\frac7{16}\zeta(3)-\frac{\pi^3}{64} ,\\
&\sum_{n=0}^{\infty} \frac{1 }
{(2n+1)^2\cosh\left ( \frac\pi2(2n+1) \right )^2 }
=\frac{\pi^2}{16}-\frac{G}{2},\\
&\sum_{n=0}^{\infty} \frac{\tanh\left ( \frac\pi2(2n+1) \right ) }
{(2n+1)\cosh\left ( \frac\pi2(2n+1) \right )^2 }
=\frac{G}{2\pi}.
\end{aligned}
Entry 4.
\begin{aligned}
&\sum_{n=0}^{\infty} \frac{\coth\left ( \frac\pi2(2n+1) \right ) }{(2n+1)^3}
=-\frac{\pi}{8}S
-\frac{\pi^3}{96}+\frac{3\pi^2}4\ln(2)-\frac{\pi G}2,\\
&\sum_{n=0}^{\infty}\frac{1}{(2n+1)^3\left ( e^{\pi(2n+1)}-1 \right ) }
=-\frac7{16}\zeta(3)-\frac{\pi}{16}S
-\frac{\pi^3}{192}+\frac{3\pi^2}8\ln(2)-\frac{\pi G}4,\\
&\sum_{n=0}^{\infty} \frac{1 }
{(2n+1)^2\sinh\left ( \frac\pi2(2n+1) \right )^2 }
=\frac14S +\frac{\pi^2}{48}-\frac{3\pi}2\ln(2)+2G,\\
&\sum_{n=0}^{\infty} \frac{\coth\left ( \frac\pi2(2n+1) \right ) }
{(2n+1)\sinh\left ( \frac\pi2(2n+1) \right )^2 }
=\frac34\ln(2)-\frac{G}\pi-\frac1{64}\,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ).
\end{aligned}
Entry 5.
$$
\displaystyle
\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3\cosh\left ( \frac\pi2
(2n+1) \right ) }
=\frac{\pi^3}{32}+\frac{\pi^2}{8}-\frac{\Gamma\left ( \frac14 \right )^4 }{16\pi}
+\frac{\pi^2}{8}\sum_{n=0}^{\infty}
\frac{\left ( \frac12 \right )_n^3 }{(1)_n^3} \frac{1}{2n+1}+\frac{\Gamma\left ( \frac14 \right )^4}{64}\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^2}{(1)_n^2}
\frac{1}{(4n+3)^2},
$$
$$
\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)\sinh\left ( \frac\pi2(2n+1) \right )^2 }
=\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\Gamma\left ( \frac{n}2 \right )
\Gamma\left ( \frac{n+1}{2} \right ) }{\Gamma\left ( \frac{n}2+\frac34 \right )^2 }.
$$
Entry 6.
\begin{aligned}
&\sum_{n=0}^{\infty}\frac{1}{2n+1}
\frac{\sinh\left ( \pi\left ( n+\frac12 \right ) \frac{K^\prime}{K} \right ) }{
\cosh\left ( 2\pi\left ( n+\frac12 \right ) \frac{K^\prime}{K} \right )}
=\frac{1}{2}\operatorname{artanh} \left (
\sqrt{\frac{1-\sqrt{1-k^2} }{2} } \right ),\\
&\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}
\frac{\cosh\left ( \pi\left ( n+\frac12 \right ) \frac{K^\prime}{K} \right ) }{
\cosh\left ( 2\pi\left ( n+\frac12 \right ) \frac{K^\prime}{K} \right )}
=\frac{1}{2}\operatorname{arctan} \left (
\sqrt{\frac{1/\sqrt{1-k^2}-1 }{2} } \right ),\\
&\sum_{m_i\in\mathbb{Z}}
\frac{1}{\cosh\left ( \pi\sqrt{m_1^2+m_2^2+m_3^2} \right ) }
=\frac{\Gamma\left ( \frac14 \right )^4 }{2\sqrt{2}\pi^3 },\\
&\sum_{m,n\ge0} \frac{(-1)^m}{\sqrt{(2m+1)(2n+1)}
\sinh\left ( \frac\pi2\sqrt{(2m+1)(2n+1)} \right ) }
=\frac12,\\
&{}_9F_8\left ( \left \{ \frac12 \right \}_8,\frac54 ;\frac14,\left \{ 1\right \} _7;1 \right )
=\frac{1024}{3\pi} \sum_{m_i\ge0}\frac{(-1)^{m_1+m_2+m_3}\left ( m_1+\frac12 \right )(m_2+\frac12)(m_3+\frac12) }{
\cosh\left ( \pi\sqrt{\left ( m_1+\frac12 \right )^2
+\left ( m_2+\frac12 \right )^2+\left ( m_3+\frac12 \right )^2 } \right ) }.
\end{aligned}
Entry 7.
\begin{aligned}
&\sum_{n = 1}^{\infty} \frac{1}{n^3\sinh(\pi n)} = -\frac{\pi}{8}S-\frac{2\pi^3}{45}+\frac{3\pi^2}{4}\ln(2)
-\frac{\pi G}{2},\\
&\sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^2\sinh(\pi n)}
=\frac{1}{8}S+\frac{23\pi^2}{360}-\frac{3\pi}{4}\ln(2)
+\frac{5 G}{12},\\
&\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3\sinh(\pi n)} =-\frac{\pi^3}{360},\\
&\sum_{n=1}^{\infty} \frac{(-1)^n\coth(\pi n)}{n^2\sinh(\pi n)}
=\frac{\pi^2}{45}-\frac{G}{3},\\
&\sum_{n=1}^{\infty} \frac{(-1)^n}{n\sinh(\pi n)^3}
=-\frac{17\pi}{240}+\frac{G}{6\pi}+\frac{1}{4}\ln(2).
\end{aligned}
Entry 8. 3 remarkable integrals:
$$
\int_{0}^{1} \frac{K^\prime{}^2}{\sqrt{k}K }\text{d}k
=\int_{0}^{1} \frac{K\left ( \sqrt{1-k^2} \right ) ^2}{\sqrt{k}K(k) }\text{d}k
=3\pi^2\ln(2)-\frac{\pi^2}{16} \,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ),
$$
$$
\int_{0}^{1} \frac{K^\prime{}^2}{K}\frac1{\sqrt{1-k^2} }\text{d}k
=\frac{3\pi^2}4\ln(2)-\frac{\pi^2}{64} \,_5F_4\left ( 1,1,\frac32,\frac32,\frac32;2,2,2,2;1 \right ),
$$
$$\int_{0}^{1} \frac{K^\prime}{K}\frac{\text{d}k}{\sqrt{k}(1-k^2)^{3/4}}
=\frac{5\pi}{2}\ln(2)-\frac{\pi}{16}
\,_4F_3\left ( 1,1,\frac32,\frac32;2,2,2;\frac12 \right ),$$
$$
\int_{0}^{1} \frac{K^\prime}{K}
\frac{\text{d}k}{(1-k^2)^{3/4}}
=\frac{\pi}{\sqrt{2} }\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ),
$$
$$
\int_{0}^{1} \frac{K^\prime}{K}
\frac{\text{d}k}{\sqrt{k}\sqrt{1-k^2}}
=\sqrt{2}\pi\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;\frac12 \right ),
$$and a hypergeometric identity:
$$
\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^5 }{(1)_n^5}
(4n+1)
=\frac{2\sqrt{2}\,\Gamma\left ( \frac14 \right )^2 }{\pi^{5/2}}
\,_4F_3\left ( \frac14,\frac14,\frac14,\frac14;\frac12,\frac12,1;1 \right )
-\frac{2}{\pi}\,_4F_3\left ( \frac14,\frac12,\frac12,\frac12;\frac34,1,1;1 \right )
$$
with a generalization, for any $\Re(s)>0$,
$$
\,_7F_6\left ( \left \{ \frac12 \right \}_5,\frac{5}{4},1-s;
s+\frac12,\frac14,\left \{ 1 \right \}_4;1 \right )
=4^{1-s}
\frac{\Gamma\left ( \frac14 \right )^2\Gamma(2s) }{\pi^2\,\Gamma\left ( s+\frac14 \right )^2 }
\,_4F_3 \left ( \frac14,\frac14,s,s;s+\frac14,s+\frac14,1;1 \right )
-\frac{2}{\pi}\,_4F_3\left ( \left \{ \frac12 \right \}_3,s;s+\frac12,1,1;1 \right ).
$$
Entry 9. An integral derived by a hyperbolic series(with a nasty looking):
$$
\int_{0}^{\frac{\sqrt{6}-\sqrt{2} }{4} }
\frac{(1-2k^2)(1-136k^2+136k^4)}{\sqrt{1-k^2} }
\left ( K^\prime-\sqrt{3}K \right ) ^6\text{d}k
=\frac{\pi^7}{16}.
$$
Entry 10. We involve some hypergeometric cases.
$$
\sum_{n=0}^{\infty} \frac{\tanh\left ( \pi\left ( n+\frac12 \right )
\frac{K^\prime}{K} \right ) }{(2n+1)\cosh\left ( \pi\left ( n+\frac12 \right )
\frac{K^\prime}{K}\right ) }
=\frac{k}2\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;k^2 \right )
,$$
$$
\sum_{n=1}^{\infty} \frac{(-1)^n\tanh(\pi n)}{n\cosh(\pi n)}
=\sqrt{2}\,_3F_2\left ( \frac12,\frac12,\frac12;1,\frac32;1 \right ) -\frac\pi2.
$$
