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The sum can be rewritten as an integral, by means of $$\frac{1}{\sin (\pi j/n)}=\frac{n}{\pi}\int_0^\infty\frac{s^{j-1}}{ s^n+1}\,ds$$ $$\Rightarrow \sum_{j=1}^{n-1}\frac{1}{\sin (\pi j/n)}=-\frac{n}{\pi}\int_0^\infty \frac{(-s)^n+s}{s (s+1)(s^n+1)}\,ds.$$ The integral does not seem to have a closed-form answer for arbitrary (even) $n$.