Not a complete answer.

Denote $n=N/2$. Note that $$ \sum _{m=1}^n \cos ^{2 s+1}\left(\frac{2 \pi m}{2 n+1}\right)=-\frac{1}{2},\qquad s=0,1,\ldots,n-1. $$ This is an elementary identity, that can be checked independently. Or one can just observe that it is a consequence of Chebyshev-Gauss quadrature.

Now, expanding the sine into Taylor series and summation with the above formula gives for the leading non-zero term $$ S(a,N)\approx (-1)^{N/2}\frac{4(2 a)^{N-1}}{N!}\left(\frac{1}{2}+\sum_{m=1}^n\cos ^{N+1}\left(\frac{2 \pi m}{N+1}\right)\right). $$ This approximate formula has been confirmed numerically: With increasing $N$, the ratio of the sum and the approximate expression seems to approach $1$. Thus, the approximate expression captures the dependence on $a$. However what to do next is unclear. The sum in paranthesis is also exponentially small. And how to prove that the terms with higher powers of $a$ can be neglected?

First, we rewrite the sum as a sum over the full period $$ S(a,N)=\frac{2}{N+1}\sum_{j=1}^{N+1} \sin^2\left( \frac{2\pi j}{N+1} \right)\sin \left( 2 a \cos \left( \frac{2\pi j}{N+1} \right) \right). $$ Denote $$ g(k,m)=\sum _{j=1}^m \cos ^{k}\left(\frac{2 \pi j}{m}\right). $$ Then the coefficients $a^{2k+1}$ of the Taylor series expansion of $S(a,N)$ is $$ \frac{2(-1)^k}{N+1}\frac{2^{2k+1}}{(2k+1)!}\left(g(2k+1,N+1)-g(2k+3,N+1)\right). $$ It is known that (see e.g. this article ) $$ g(k,m)=\frac{m}{2^k}\sum_{r=-\lfloor k/m\rfloor,\,rm+k\,\mathrm{even}}^{\lfloor k/m\rfloor}\binom{k}{\tfrac{rm+k}{2}}. $$ In particular $$ g(2k+1,2n+1)=0,\qquad k=0,1,2,\ldots,n-1.\quad (n=N/2) $$ Observe that this is also a consequence of Chebyshev-Gauss quadrature.

Now, summation with the above formula shows that coefficient of $a^{2k+1}$ in the Taylor series expansion of $S(a,N)$ with $k=0,1,2,\ldots,n-2$ vanish. The first non-vanishing term corresponds to $k=n-1$: \begin{align} S(a,N)&\approx (-1)^{N/2}\frac{2(2 a)^{N-1}}{N!}g(N+1,N+1)\\ &=(-1)^{N/2}\frac{2(2 a)^{N-1}}{N!}\frac{2(N+1)}{2^{N+1}}\\ &\approx \frac{(-1)^{N/2} a^{N-1}}{(N-1)!}. \end{align} This approximate formula has been confirmed numerically: With increasing $N$, the ratio of the sum and the approximate expression seems to approach $1$.

The higher order coefficients probably can be estimated too with the formulas above, and they will be much smaller than the leading coefficient. This is left as an exercise.