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, $$\lim_{p\rightarrow\infty}\frac{1}{2p}\sum_{j=1}^{2p-1}\frac{(-1)^j}{\sin (\pi j/n)}= -\frac{2}{\pi}\int_0^1\frac{1}{s+1}\,ds=-\frac{2}{\pi}\ln 2=-0.441271$$
Higher order terms in a series expansion were evaluated in On a generalization of Watson's trigonometric sum, or Dowker's sum of order one half, $$\sum_{j=1}^{n-1}\frac{(-1)^j}{\sin (\pi j/n)} = -\frac{\,2n\ln2\,}{\,\pi\,}\: - 2\sum_{r=1}^{N-1} \frac{\,(-1)^{r+1}\big(2^{2r-1}-1\big) \big(2^{2r}-1\big)\pi^{2r-1} B^2_{2r}\,}{\,r\, (2r)!\, n^{2r-1}\,} \, + \, O\big(n^{1-2N}\big),$$ where the $B_p$ are Bernoulli numbers.
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 On a finite sum of cosecants appearing in various problems
