$\DeclareMathOperator{\Res}{Res}$
Partial answer: closed form of the sum
$$S_m(n)=\sum_{k=1}^{n-1}\csc^{2m}\left(\frac{k\pi}{n}\right)$$
$$=\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{1}{2p+1}\sum_{s=0}^{p}\left(\left[z^{-(2s+1)}\right]\cot^{2p+1}(z)\right)\frac{(2s+1)2^{2s+2}|B_{2n+2}|}{(2s+2)!}(n^{2s+2}-1)$$
Added: Alright, after all those calculations, while looking at some related question I bumped into this answer by Jack D'Aurizio, which completely overshadowed my answer, so shout out to him as well.
Proof (I omitted some obvious algebra steps, otherwise the proof would be too long, meanwhile, I'm not sure how I can start proving OP's conjecture with this closed form)
Start with
$$\csc^{2m}(x)=\csc^2(x)(1+\cot^2(x))^{m-1}=-\left(\frac{d}{dx}\cot(x)\right)\sum_{p=0}^{m-1}\binom{m-1}{p}\cot^{2p}(x)$$
$$\implies\csc^{2l}(\pi x)=\frac{d}{dx}\left(\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{-1/\pi}{2p+1}\cot^{2p+1}(\pi x)\right)$$
$$=\frac{1}{\pi}\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{1}{2p+1}\underset{z=x}{\Res}\frac{-\cot^{2p+1}(\pi z)}{(z-x)^2}\ ;x\not\in\mathbb{Z}$$
Using residue theorem:
$$\underset{z=x}{\Res}\frac{\cot^{2p+1}(\pi z)}{(z-x)^2}+\sum_{l\in\mathbb{Z}}\underset{z=l}{\Res}\frac{\cot^{2p+1}(\pi z)}{(z-x)^2}=\lim_{N\to\infty}\frac{1}{2\pi i}\oint_{\Gamma_N}\frac{\cot^{2p+1}(\pi z)}{(z-x)^2}dz$$
The integral in the limit on RHS is $O(N^{-2})$ since $\left|\cot^{2p+1}(\pi z)\right|$ is bounded (see this answer)
Thus we have
$$S_m(n)=\sum_{k=1}^{n-1}\csc^{2m}\left(\pi\frac{k}{n}\right)$$ $$=\frac{1}{\pi}\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{1}{2p+1}\sum_{l\in\mathbb{Z}}\underset{z=l}{\Res}\left(\cot^{2p+1}(\pi z)\sum_{k=1}^{n-1}\frac{1}{\left(z-\frac{k}{n}\right)^2}\right)$$
Some algebra manipulation:
$$\sum_{l\in\mathbb{Z}}\underset{z=l}{\Res}\left(\cot^{2p+1}(\pi z)\sum_{k=1}^{n-1}\frac{1}{\left(z-\frac{k}{n}\right)^2}\right)=\underset{z=0}{\Res}\left(\cot^{2p+1}(\pi z)\sum_{k=1}^{n-1}\sum_{l\in\mathbb{Z}}\frac{1}{\left(z+l-\frac{k}{n}\right)^2}\right)$$
$$=\underset{z=0}{\Res}\left(\cot^{2p+1}(\pi z)\sum_{k=1}^{n-1}\pi^2\csc^{2}\left(\pi\frac{k}{n}-\pi z\right)\right)$$
Plug that back in our formula of $S_m(n)$
$$S_m(n)=\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{1}{2p+1}\left[z^{-1}\right]\pi\cot^{2p+1}(\pi z)\sum_{k=1}^{n-1}\csc^{2}\left(\pi\frac{k}{n}-\pi z\right)$$
Consider the sum $\sum_{k=1}^{n-1}\csc^{2}\left(\pi\frac{k}{n}-\pi z\right)$, we use trigamma function to obtain its closed form
$$\sum_{k=1}^{n-1}\csc^{2}\left(\pi\frac{k}{n}-\pi z\right)=\frac{1}{\pi^2}\sum_{k=1}^{n-1}\left(\psi^{(1)}\left(\frac{k}{n}-z\right)+\psi^{(1)}\left(1-\frac{k}{n}+z\right)\right)$$
$$=\frac{1}{\pi^2}\sum_{k=1}^{n-1}\left(\psi^{(1)}\left(\frac{k}{n}-z\right)+\psi^{(1)}\left(\frac{k}{n}+z\right)\right)$$
$$=\frac{n^2}{\pi^2}\sum_{q=0}^{\infty}\sum_{k=1}^{n-1}\left(\frac{1}{(k+(q-z)n)^2}+\frac{1}{(k+(q+z)n)^2}\right)$$
$$=\frac{n^2}{\pi^2}\sum_{q=0}^{\infty}\left(\psi^{(1)}(1+(q-z)n)-\psi^{(1)}((q-z+1)n)+\psi^{(1)}(1+(q+z)n)-\psi^{(1)}((q+z+1)n)\right)$$
Use the fact that $\psi^{(1)}(z)=\psi^{(1)}(z+1)+\frac{1}{z^2}$, we get a nice telescoping sum, and the result is
$$\frac{1}{\pi^2}\left(n^2\psi^{(1)}(1-nz)+n^2\psi^{(1)}(1+nz)-\psi^{(1)}(1-z)-\psi^{(1)}(1+z)\right)$$
Use Taylor expansion, we get
$$\frac{1}{\pi^2}\left(\psi^{(1)}(1-z)+\psi^{(1)}(1+z)\right)=\frac{1}{\pi^2}\sum_{s=0}^{\infty}(2s+1)\zeta(2(s+1))z^{2s}$$
$$=\sum_{s=0}^{\infty}\frac{(2s+1)2^{2s+1}|B_{2n+2}|}{(2s+2)!}(\pi z)^{2s}$$
So
$$\sum_{k=1}^{n-1}\csc^{2}\left(\pi\frac{k}{n}-\pi z\right)=\sum_{s=0}^{\infty}\frac{(n^{2s+2}-1)(2s+1)2^{2s+2}|B_{2n+2}|}{(2s+2)!}(\pi z)^{2s}$$
Putting everything together:
$$S_m(n)=\sum_{p=0}^{m-1}\binom{m-1}{p}\frac{1}{2p+1}\sum_{s=0}^{p}\frac{c_{-(2s+1)}(2s+1)2^{2s+2}|B_{2n+2}|}{(2s+2)!}(n^{2s+2}-1)$$
Where $c_n=[z^n]\cot^{2p+1}(z)$
