I'm not sure if this representation is really helpful, but the above sum $S_n$ can be written in terms of the $q$-digamma function $\psi_q(z)$. Define $q=r\exp(2\pi i/n)$ and $m=\lfloor n/4\rfloor$, with $r<1$ and even $n$, then \begin{align} S_n &= 1 + \frac{n}{\pi}\lim_{r\to1^-} \big[ \psi_q\left( 1 \right) -\psi_q\left(\tfrac{1}{2}\right) +\psi_q\left(m+\tfrac{n}{2}+1\right) -\psi_q\left(m+\tfrac{n}{2}+\tfrac{1}{2}\right) \\ & \qquad{} -\psi_q(m+1) +\psi_q\left(m+\tfrac{1}{2}\right) -\psi_q\left(\tfrac{n}{2}+1\right) +\psi_q\left(\tfrac{n}{2}+\tfrac{1}{2}\right) \big]. \label{eq:1}\tag{1} \end{align}

Mathematica finds this representation if one splits the sum into even and odd parts.

I'm not sure if this representation is really helpful, but the above sum $S_n$ can be written in terms of the $q$-digamma function $\psi_q(z)$. Define $q=r\exp(2\pi i/n)$ and $m=\lfloor n/4\rfloor$, with $r<1$ and even $n$, then \begin{align} S_n &= 1 + \frac{n}{\pi}\lim_{r\to1^-} \big[ \psi_q\left( 1 \right) -\psi_q\left(\tfrac{1}{2}\right) +\psi_q\left(m+\tfrac{n}{2}+1\right) -\psi_q\left(m+\tfrac{n}{2}+\tfrac{1}{2}\right) \\ & \qquad{} -\psi_q(m+1) +\psi_q\left(m+\tfrac{1}{2}\right) -\psi_q\left(\tfrac{n}{2}+1\right) +\psi_q\left(\tfrac{n}{2}+\tfrac{1}{2}\right) \big]. \label{eq:1}\tag{1} \end{align}

Mathematica finds this representation if one uses the symmetry $ j \leftrightarrow n-j$ and splits the sum into even and odd parts.