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) $n$.

The large-$n$ asymptotics for even $n=2p$ can be readily derived from this integral expression [split the integration interval into $\int_0^1$ and $\int_1^\infty$, both intervals contribute $-(n/\pi)\ln 2$], with the result
$$\lim_{p\rightarrow\infty}\frac{1}{2p}\sum_{j=1}^{2p-1}\frac{1}{\sin (\pi j/n)}=-\frac{2}{\pi}\ln 2.$$