I applied residue theorem to the function $f(z) = \frac{\cot z \coth z}{z^3}$ on box contour $\Gamma_N$ that has corners, $\pm \pi(N+1/2) \pm i \pi (N+1/2)$.
I obtain, residues $Res(f,n\pi) = \frac{\coth n \pi }{n^3 \pi ^3}$ where $n\not=0$, and $Res(f,0)= \frac{-7}{45}$. By residue theorem, as $\int_{\Gamma_N} f \rightarrow 0 $ as $N \rightarrow \infty$.
$$ 2 \sum_{n=1}^\infty \frac{\coth n \pi } {n^3 \pi^3} = \frac{7}{45} \Rightarrow \sum_{n=1}^\infty \frac{\coth n \pi }{n^3} = \frac{7 \pi^3}{90}.$$
But the answer I searched online is $\frac{7 \pi^3}{180}$. What could have possibly gone wrong?
You have considered all the poles which make $\cot z$ singular. These are all real. Let $z=i y$. $$\cot z \coth z=\frac{\cos(iy)\cosh(iy)}{\sin(iy)\sinh(iy)}=\frac{\cosh y\cos y}{i\sinh y\cdot(-i)\sin y}=\cot y\coth y$$ So any poles you had for $z=x\in\Bbb R$ are also poles for $z=iy, y\in\Bbb R$, and have the same residues. So you'll end up with twice the sum which fixes the factor of 2.