This question and this question are about $$\sum_{n=1}^\infty \frac{1}{\cosh n\pi}=\frac{1}{2}\left(\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}-1\right)$$ and $$\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}=\frac{1}{6}-\frac{1}{2\pi}$$ But I couldn't find the closed form (in terms of $\Gamma (a)$ for rational $a$) of $$S=\sum_{n=1}^\infty \frac{1}{\sinh n\pi}.$$
The series $S$ can be rewritten as Lambert series $$2\sum_{n=1}^\infty \frac{q^n}{1-q^{2n}}$$ where $q=e^{-\pi}$, but unlike $$\sum_{n=1}^\infty \frac{1}{\cosh n\pi}=2\sum_{n=1}^\infty \frac{q^n}{1+q^{2n}}=2\sum_{n=0}^\infty (-1)^n \frac{q^{2n+1}}{1-q^{2n+1}},$$ $S$ doesn't seem to be expressible in terms of $\vartheta_3(x)$ (Jacobi theta function).
The other series $\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}$ is suggestive of the Weierstrass elliptic function $\wp$: $$\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}=\frac{1}{2\pi ^2}\sum_{n=1}^\infty \sum_{m\in\mathbb{Z}}\left(\frac{1}{(n-mi)^2}+\frac{1}{(n+mi)^2}\right)$$ but this approach is exploiting the square of $\sinh$ in the denominator which isn't of any use for $S$.
Mathematica gives the following result where $\psi _{q}^{(0)}(z)$ is the q-Polygamma Function.
$$\sum _{n=1}^{\infty } \frac{1}{\sinh (\pi n)}=\frac{\psi _{e^{\pi }}^{(0)}(1-i)-\psi _{e^{\pi }}^{(0)}(1)+i \pi }{\pi }$$