Your final integral can be readily evaluated by expanding the fraction of sines into sums of exponentials $e^{ikx/2}$ with integer $k$, and integrating term by term with the Gaussian weight, to arrive at $$|L(0)|^2 \equiv\frac{M}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} \frac{\sin^2\left(Nx/2 \right)}{\sin^2\left( x/2 \right)} \, dx=$$ $$=MN+2M e^{-N^2} \sum _{k=1}^{N-1} k e^{\frac{1}{2} \left(N^2-k^2+2 k N\right)}.$$

Your final integral can be readily evaluated by expanding the fraction of sines into sums of exponentials $e^{ikx/2}$ with integer $k$, and integrating term by term with the Gaussian weight, to arrive at $$|L(0)|^2 \equiv\frac{M}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} \frac{\sin^2\left(Nx/2 \right)}{\sin^2\left( x/2 \right)} \, dx=$$ $$=MN+2M e^{-N^2/2} \sum _{k=1}^{N-1} k e^{\frac{1}{2} \left(2 k N-k^2\right)}.$$