Letting $\xi=e^{\frac{2\pi i}{2n}}$, a primitive root of unity $z^{2n}=1$. Then, $\sin\left(\frac{\pi k}n\right)=\frac{\xi^{k}-\xi^{-k}}{2i}$ and hence
\begin{align}
\sum_{k=0}^{n-1}\sin^m\left(\frac{\pi k}n\right)
&=\frac1{(2i)^m}\sum_{k=0}^{n-1}\sum_{j=0}^m\binom{m}j(-1)^j\xi^{-jk}\xi^{(m-j)k} \\
&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\sum_{k=0}^{n-1}\xi^{(m-2j)k} \\
&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{(m-2j)n}-1}{\xi^{m-2j}-1} \\
&=\frac1{(2i)^m}\sum_{j=0}^m(-1)^j\binom{m}j\frac{\xi^{mn}-1}{\xi^{m-2j}-1}.
\end{align}
The outcome is *parity-dependent*. For example, if $m\rightarrow 2m$ is even then the average value is
$$\frac1n\sum_{k=0}^{n-1}\sin^{2m}\left(\frac{\pi k}n\right)
=\frac1{2^{2m}}\binom{2m}m,$$
as long as $n>m$. This is because $\xi^{2mn}-1=0$ and the only time $\xi^{2m-2j}-1=0$ is when $j=m$ (since $n>m$). Therefore,
$$\sum_{k=0}^{n-1}\sin^{2m}\left(\frac{\pi k}n\right)
=\frac1{(2i)^{2m}}\cdot (-1)^m\binom{2m}m=\frac1{2^{2m}}\binom{2m}m.$$
By the way, 
$$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$
I believe a similar computation can be worked out for $m$ odd.