The sum can be rewritten as an integral, by means of
$$\frac{1}{\sin (\pi j/n)}=\frac{n}{\pi}\int_0^\infty\frac{s^{j-1}}{  s^n+1}\,ds,\;\;0<j<n,$$
$$\Rightarrow \sum_{j=1}^{n-1}\frac{(-1)^j}{\sin (\pi j/n)}=-\frac{n}{\pi}\int_0^\infty \frac{(-s)^{n-1}+1}{ (s+1)(s^n+1)}\,ds$$
$$\qquad=\begin{cases}
0&\text{for odd}\;\;n,\\
-\frac{2n}{\pi}\int_0^1\frac{s^{n-1}+1}{ (s+1)(s^n+1)}\,ds&\text{for even}\;\;n.
\end{cases}
$$
This does not simplify further for arbitrary even $n$.

The large-$n$ asymptotics for even $n=2p$ can be readily derived from the integral expression,
$$\lim_{p\rightarrow\infty}\frac{1}{2p}\sum_{j=1}^{2p-1}\frac{(-1)^j}{\sin (\pi j/n)}= -\frac{2}{\pi}\int_0^1\frac{1}{s+1}\,ds=-\frac{2}{\pi}\ln 2=-0.441271$$

Here is a plot of $\frac{1}{2p}\sum_{j=1}^{2p-1}\frac{(-1)^j}{\sin (\pi j/n)}$ as a function of $p$.

<IMG SRC="https://i.sstatic.net/82QOJ4lT.png" WIDTH="400"/>


  [1]: https://i.sstatic.net/82QOJ4lT.png