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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, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed formula in the case $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: $$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\right).$$

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, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed formula in the case $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: $$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\pi\right).$$

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, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed form when $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: $$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\right).$$

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, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed formula in the case $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: $$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\right).$$

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.

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, $$\frac1{\pi}\int_0^{\pi}\sin^{2m}x\,dx=\frac1{2^{2m}}\binom{2m}m.$$ Contrary to what I thought initially, things are a bit tricky (or unlikely) that there is a closed form when $m\rightarrow 2m+1$ is odd. What we get with the above approach is just an identity: $$\sum_{k=0}^{n-1}\sin^{2m+1}\left(\frac{\pi k}n\right) =\frac1{2^{2m}}\sum_{j=0}^m(-1)^j\binom{2m+1}{m-j}\cot\left(\frac{2j+1}{2n}\right).$$