I believe the problem can be approached from the generating-functions perspective.

First we notice that at least two of the $k_i$'s are positive. Based on the symmetry, let's consider the sum with $k_1,k_2>0$ and multiply it by 3. This will cover all cases, but the case with all $k_i>0$ will be counted thrice. Hence,
$$S_N = 3S_N^{(2)} - 2S_N^{(1)},$$
where in $S_N^{(2)}$ sums under the assumption $k_1,k_2>0$, while $S_N^{(1)}$ sums under the assumption $k_1,k_2,k_3>0$.

Let me consider $S_N^{(1)}$, which takes the form:
$$S_N^{(1)} = (\sin\alpha)^3\sum_{k_1+k_2+k_3=N\atop k_1,k_2,k_3>0} \frac{1}{k_1k_2k_3} U_{k_1-1}(\cos\alpha)U_{k_2-1}(\cos\alpha)U_{k_3-1}(\cos\alpha),$$
where $U_k()$ are Chebyshev polynomials of second kind and $\alpha:=\frac{\pi}{N}$.

From the generating function for Chebyshev polynomials we have
$$\sum_{m\geq 1} \frac{U_{m-1}(\cos\alpha)}m t^m =\frac{\arctan\frac{t\sin\alpha}{1-t\cos\alpha}}{\sin\alpha}.$$
Hence,
$$S_N^{(1)} = [t^N]\ (\arctan\frac{t\sin\alpha}{1-t\cos\alpha})^3.$$
So, essentially we got the generating function for $S_N^{(1)}$. It should be possible to get the asymptotic by standard means (e.g., see Section 5.4 in generatingfunctionology).

**UPDATE.** Here is yet another (more straightforward) take on the original problem.

Let $z:=e^{I\frac{\pi}{N}}$ and notice that $\sin(k\alpha) = -\frac{I}{2}(z^k - z^{-k})$, where $I$ is the imaginary unit. Taking into account symmetry in summation indices and that $z^N=-1$, we get
\begin{split}
S_N &= \frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{k_1}-z^{-k_1}) (z^{k_2}-z^{-k_2}) (z^{k_3}-z^{-k_3})\\
&=\frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} \big( (z^{k_1+k_2+k_3}-z^{-(k_1+k_2+k_3)}) - 3(z^{k_1+k_2-k_3}-z^{-(k_1+k_2-k_3)})\big)\\
&=-\frac{3I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{2k_3}-z^{-2k_3}).
\end{split}

We will need generating functions:
$$\sum_{k=1}^{N} \frac{t^k}{k} = \int_0^t dx \frac{1-x^N}{1-x},$$
$$\sum_{k=-(N-2)}^{-1} \frac{t^k}{k} = \sum_{k=2}^{N-1}\frac{t^{k-N}}{k-N} = \int_0^t dx\frac{1-x^{N-2}}{(1-x)x^{N-1}},$$
and sum of the two:
$$F_N(t) := \sum_{-(N-2)\leq k\leq N\atop k\ne 0} \frac{t^k}{k} = \int_0^t dx \frac{1-x^{N-2}+x^{N-1}-x^{2N-1}}{(1-x)x^{N-1}} $$

Then
$$S_N = -\frac{3I}{8} [t^N]\ F_N(t)^2\big( F_N(z^2t) - F_N(z^{-2}t)\big)$$