I'll try to summarize as much as possible, and show you what I've done. $$ \Omega = \sum_{n=1}^{\infty } \frac{\tan^{-1}(n)}{n^{2}} = \frac{\pi^{3}}{12} - \sum_{n=1}^{\infty } \frac{\tan^{-1}(\frac{1}{n})}{n^{2}} = \frac{\pi^{3}}{12} - I$$ $$ I = \sum_{n=1}^{\infty } \frac{\tan^{-1}(\frac{1}{n})}{n^{2}} = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1}\zeta (2k+3) = \int_{0}^{\infty}\frac{1}{e^{u}-1}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)(2k+2)!}u^{2k+2} $$ $$ = \int_{0}^{\infty} \frac{cos(x)-1}{e^{x}-1} dx +\int_{0}^{\infty} \frac{xSi(x)}{e^{x}-1} dx = A + B $$ $$ A = \int_{0}^{\infty} \frac{cos(x)-1}{e^{x}-1} dx = \frac{1}{2}\int_{0}^{1}\frac{t^{i} + t^{-i} -2}{1-t}dt = -\gamma -\frac{1}{2}(\psi(1+i) + \psi(1-i)) \\ = -\gamma - \Re(\psi(i)) = \sum_{n=0}^{\infty}\frac{n-1}{n^{3}+n^{2}+n+1} $$ $$ B = \int_{0}^{\infty} \frac{xSi(x)}{e^{x}-1} dx = \sum_{k=1}^{\infty} \frac{(-1)^{k}2k}{2k-1}\zeta (2k+1) $$
This is where I stopped,