The sum can be rewritten as an integral, by means of
$$\frac{1}{\sin (\pi j/n)}=\frac{n}{\pi}\int_0^\infty\frac{s^{j-1}}{  s^n+1}\,ds,\;\;0<j<n,$$
$$\Rightarrow \sum_{j=1}^{n-1}\frac{(-1)^j}{\sin (\pi j/n)}=-\frac{n}{\pi}\int_0^\infty \frac{(-s)^{n-1}+1}{ (s+1)(s^n+1)}\,ds=\begin{cases}
0&\text{for odd}\;\;n,\\
-\frac{2n}{\pi}\int_0^1\frac{s^{n-1}+1}{ (s+1)(s^n+1)}\,ds&\text{for even}\;\;n.
\end{cases}
$$
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,
$$\lim_{p\rightarrow\infty}\frac{1}{2p}\sum_{j=1}^{2p-1}\frac{(-1)^j}{\sin (\pi j/n)}= -\frac{2}{\pi}\int_0^1\frac{1}{s+1}\,ds=-\frac{2}{\pi}\ln 2=-0.441271$$

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As a curiosity, I note that the sum for odd $n$, with an offset $\phi$ in the argument of the sine, _can_ be evaluated in closed form: $$\sum_{j=0}^{n-1}\frac{(-1)^j}{\sin(\phi+\pi j/n)}=\frac{n}{\sin n\phi},\;\;n=1,3,5,\ldots$$
A result due to Euler, see <A HREF="https://arxiv.org/abs/2312.16657">On a finite sum of cosecants appearing in various problems</A>    

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