Let $\zeta$ be a primitive $n$-th root of unity. Then
$$\prod_{j=1}^{n-1}(x-\zeta^j)=\frac{x^n-1}{x-1}=1+x+\cdots+x^{n-1}$$
and hence
$$\sigma_k=\sum_{1\le i_1<\cdots<i_k\le n-1}\zeta^{i_1+\cdots+i_k}=(-1)^k$$
for all $k=1,\ldots,n-1$. 

Observe that
\begin{align*}\mathrm{per}[1-\zeta^{j+k}]_{1\le j,k\le n-1}=&\sum_{\tau\in S_{n-1}}\ \prod_{j=1}^{n-1}(1-\zeta^{j+\tau(j)})
\\=&\sum_{\tau\in S_{n-1}}1+\sum_{\tau\in S_{n-1}}\ \sum_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum_{j\in J}\ (j+\tau(j))}
\\=&(n-1)!+\sum_{\emptyset \not=J\subseteq \{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum_{j\in J}j}\sum_{\tau\in S_{n-1}}\zeta^{\sum_{j\in J}\ \tau(j)}.
\end{align*}
For $\emptyset \not=J\subseteq\{1,\ldots,n-1\}$, clearly
\begin{align*}&\sum_{\tau\in S_{n-1}}\zeta^{\sum_{j\in J}\ \tau(j)}
\\=&\sum_{1\le i_1<\cdots<i_{|J|}\le n-1}\zeta^{i_1+\cdots+i_{|J|}}\sum_{\tau\in S_{n-1}\atop\{\tau(j):\ j\in J\}
=\{i_1,\ldots,i_{|J|}\}}1
\\=&|J|!(n-1-|J|)!\sigma_{|J|}=(-1)^{|J|}|J|!(n-1-|J|)!.
\end{align*}
Therefore
\begin{align*}\mathrm{per}[1-\zeta^{j+k}]_{1\le j,k\le n-1}=&(n-1)!+\sum_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}|J|!(n-1-|J|)!\zeta^{\sum_{j\in J}\ j}
\\=&(n-1)!\sum_{k=0}^{n-1}\frac{(-1)^k}{\binom{n-1}k}=(1-(-1)^n)\frac{n!}{n+1}.
\end{align*}
Then the conjectural identity $(2)$ follows from this since
$$\sin\pi\frac{j+k}n=\frac{e^{-\pi i(j+k)/n}}{2i}\left(e^{2\pi i(j+k)/n}-1\right).$$

The identities $(3)$ can be proved similarly, in fact we have
$$\mathrm{per}[1+\zeta^{j+k}x]_{1\le j,k\le n-1}=(n-1)!\sum_{k=0}^{n-1}\frac{x^k}{\binom{n-1}k}.$$ The idea here is slight modification of my way to establish Theorem 1.1 in my preprint *Arithmetic properties of some permanents* available from http://arxiv.org/abs/2108.07723.

I admit that the identities $(1)$ and $(4)$ remain open.