I wrote my try below. It does not provide an answer and so I did not checked the correctness of all operations. Nevertheless, I hope it can help to decide to follow this way or not.
In order to have the sum in the expression for $S_n$ be nonempty, we assume that $n\ge 2$.
Put $n'=\left\lfloor \frac n2\right\rfloor$ and $\xi=e^{\pi i/n}$.
Then $\xi^{2n'}$ equals $\xi^{n}=e^{\pi i}=-1$, if $n$ is even, and equals
$\xi^{n+1}=e^{\pi i}\xi=-\xi$, if $n$ is odd.
Let $s$ be any integer and $$\Sigma(s)=\sum_{k\; {\rm odd}}^{n-1} \xi^{sk}=
\sum_{\ell=0}^{n'-1} \xi^{s(2\ell+1)}.$$
We have $\xi^s=1$ iff $2n|s$. In this case each summand of $\Sigma(s)$ is $1$ so $\Sigma(s)=n'$.
Otherwise
$$\Sigma(s)=\xi^s\sum_{\ell=0}^{n'-1} \xi^{2s\ell}=\xi^s\cdot\frac{1-\xi^{2sn'}}{1-\xi^s}=
\begin{cases}
\hskip20pt 0, & \text{if } n \text{ is even and } s \text{ is even},\\
\hskip5pt \frac {2\xi^s}{1-\xi^s}, & \text{if } n \text{ is even and } s \text{ is odd},\\
\hskip15pt \xi^s, & \text{if } n \text{ is odd and } s \text{ is even},\\
\frac {\xi^s+\xi^{2s}}{1-\xi^s}, & \text{if } n \text{ is odd and } s \text{ is odd}.\\
\end{cases}
$$
Moreover, it is easy to check that
$$\Sigma(s)+\Sigma(-s)=
\begin{cases}
\hskip63pt 0, & \text{if } n \text{ is even and } s \text{ is even},\\
\hskip55pt -2, & \text{if } n \text{ is even and } s \text{ is odd},\\
\hskip32pt \xi^s+\xi^{-s}, & \text{if } n \text{ is odd and } s \text{ is even},\\
-(2+\xi^s+\xi^{-s}), & \text{if } n \text{ is odd and } s \text{ is odd}.\\
\end{cases}
$$
Since for each real $x$ we have $\sin x=\frac{e^{xi}-e^{-xi}}{2i}$ and $\cos x=\frac{e^{xi}+e^{-xi}}2$, for each natural $k$ we have $\sin\frac{\pi k}n=\frac{\xi^k-\xi^{-k}}{2i}$, $\cos\frac{\pi k}n=\frac{\xi^k+\xi^{-k}}{2}$, and $\sin\frac{\pi kr}n=\frac{\xi^{kr}-\xi^{-rk}}{2i}$.
Suppose now that $a>0$ or $n\ne 2\pmod 4$. Put $b=\frac 1{2(a+1)}$. Then for each odd $k\le n-1$ we have $|2b\cos\frac{\pi k}n|<1$, so
$$
\frac 1{a+1-\cos\frac{\pi k}n}=2b\cdot \frac 1{1-2b\cdot \cos\frac{\pi k}n}=
$$
$$2b\cdot\sum_{m=0}^\infty (2b)^m\cdot \cos^m\frac{\pi k}n=2b\cdot\sum_{m=0}^\infty b^m\cdot \left(\xi^k+\xi^{-k}\right)^m=
$$
$$
2b\cdot\sum_{m=0}^\infty b^m\cdot \sum_{j=0}^m {m\choose j}\xi^{k(2j-m)}.
$$
Therefore
$$nS_n=4b\sum_{\ell=0}^{n'-1} \frac{\xi^{2\ell+1}-\xi^{-(2\ell+1)}}{2i}\cdot \frac{\xi^{(2\ell+1)r}-\xi^{-(2\ell+1)r}}{2i}
\times$$ $$\sum_{m=0}^\infty b^m\cdot \sum_{j=0}^m {m\choose j}\xi^{(2\ell+1)(2j-m)}=$$
$$-b\sum_{m=0}^\infty b^m \sum_{j=0}^m {m\choose j}\cdot \sum_{\ell=0}^{n'-1} (\xi^{2\ell+1}-\xi^{-(2\ell+1)})(\xi^{(2\ell+1)r}-\xi^{-(2\ell+1)r})\xi^{(2\ell+1)(2j-m)}=
$$
$$
-b\sum_{m=0}^\infty b^m P(m),$$
where $$P(m)=\sum_{j=0}^m {m\choose j}(\Sigma(1+r+2j-m)-\Sigma(1-r+2j-m)-$$ $$
\Sigma(-1+r+2j-m)+\Sigma(-1-r+2j-m)).
$$
Now there is a hope that we shall be able to cancel many summands here,
calculating the coefficients at $\Sigma(s)$ or using the expression for $\Sigma(s)+\Sigma(-s)$.
Moreover, if both $n$ and $1+r-m$ are even then all summands $\Sigma(s)$ are equal to $0$, but those with $2n|s$ which are equal to $n/2$.
Also when $b$ is small (that is, $a$ is big) maybe it suffices to calculate a $P(m)$ for a few first natural $m$ to provide a good asymptotics for $nS_n$.
Update. Unfortunately, the coefficients at $\Sigma(s)$ do not look good even for $r=1$. Namely, in this case
$$P(m)=\sum_{j=0}^m {m\choose j}(\Sigma(2+2j-m)-2\Sigma(2j-m)+\Sigma(-2+2j-m))=$$
$$\sum_{j=0}^m \Sigma(2j-m)\left({m\choose j+1}-2 {m\choose j}+{m\choose j-1}\right)=$$
$$m!\sum_{j=0}^m \Sigma(2j-m)\cdot\frac{m^2-4mj+4j^2-m-2}{(j+1)!(m-j+1)!}.$$
