Compute the Fourier Transform
$$
\begin{align}
&\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz}-2+e^{-2iz}}{-4z^2}\,e^{-2\pi iz\xi}\,\mathrm{d}z\tag1\\
&=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz(1-\pi\xi)}-2e^{-2\pi iz\xi}+e^{-2iz(1+\pi\xi)}}{-4z^2}\,\mathrm{d}z\tag2\\[6pt]
&=\pi(1-\pi\xi)[\pi\xi\le1]+2\pi(\pi\xi)[\pi\xi\lt0]-\pi(1+\pi\xi)[\pi\xi\lt-1]\tag3\\[12pt]
&=\pi(1-\pi\xi)[0\le\pi\xi\le1]+\pi(1+\pi\xi)[-1\le\pi\xi\le0]\tag4\\[12pt]
&=\pi(1-\pi|\xi|)\big[\,\pi|\xi|\le1\,\big]\tag5
\end{align}
$$
Explanation:
$(1)$: write $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\phantom{\text{(1):}}$ shift the contour since there are no singularities
$(2)$: combine exponents
$(3)$: use the contour $[-R-i,R-i]\cup Re^{i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a positive coefficient of $iz$
$\phantom{\text{(3):}}$ use the contour $[-R-i,R-i]\cup Re^{-i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a negative coefficient of $iz$
$\phantom{\text{(3):}}$ we need only count the residues from
$\phantom{\text{(3):}}$ the exponentials with a positive coefficient of $iz$
$(4)$: simplify
$(5)$: simplify
For $t\gt0$, substitute $x\mapsto x/t$ and apply $(5)$:
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(tx)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x
&=t\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi/t}\,\mathrm{d}x\tag6\\[6pt]
&=\pi(t-\pi|\xi|)\big[\,\pi|\xi|\le t\,\big]\tag7
\end{align}
$$
Apply Poisson Summation
$$
\begin{align}
t^2+2\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}
&=\sum_{k\in\mathbb{Z}}\frac{\sin^2(tk)}{k^2}\tag8\\[6pt]
&=\sum_{k\in\mathbb{Z}}\pi(t-\pi|k|)\big[\,\pi|k|\le t\,\big]\tag9\\
&=\pi t+2\sum_{k=1}^{\lfloor t/\pi\rfloor}\pi(t-\pi k)\tag{10}\\[9pt]
&=\pi t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: make a sum over $\mathbb{Z}$
$\phantom{1}(9)$: Poisson summation
$(10)$: make a sum over $\mathbb{N}$
$(11)$: sum in $k$
Solve for the sum:
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac12\left((\pi-t)t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\right)}\tag{12}
$$
A Dilogarithmic Identity
As Claude Leibovici shows using $\sin(tk)=\frac{e^{itk}-e^{-itk}}{2i}$,
$$\newcommand{\Li}{\operatorname{Li}}
\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac{\pi^2}{12}-\frac14\left(\Li_2\left(e^{2it}\right)+\Li_2\left(e^{-2it}\right)\right)\tag{13}
$$
which gives a nice identity:
$$
\hspace{-18pt}\bbox[5px,border:2px solid #C0A000]{\Li_2\left(e^{it}\right)+\Li_2\left(e^{-it}\right)=\frac{\pi^2}3-\frac12\left((2\pi-t)t+4\!\left(\pi t-\pi^2\right)\left\lfloor\frac{t}{2\pi}\right\rfloor-4\pi^2\left\lfloor\frac{t}{2\pi}\right\rfloor^2\right)}\tag{14}
$$
We do have
\begin{align*}
\int_{-\infty}^{\infty} e^{-y^2} dy &= \sqrt{\pi}
\end{align*}
Now, substitute $y = \alpha x$, $dy = \alpha dx$:
\begin{align*}
\alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx &= \sqrt{\pi} \\
\left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx \right)^2 &= \pi
\end{align*}
This suggests that we may estimate the given sums by interpreting them as a Riemann sum belonging to the integral we've just computed. However, we have to be careful.
$\sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < \int_{0}^{\infty} e^{-\alpha^2 x^2} dx$
$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 1 + 2 \sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < 1 + \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} \, dx$
And for the other bound use
$\int_0^{\infty} e^{-\alpha^2 x^2} \, dx < \sum_{n=0}^{\infty} e^{-\alpha^2 n^2}$
To find
$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 2 \sum_{n=0}^{\infty} e^{-\alpha^2 n^2} - 1 > \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx$
Collect both halves and multiply with
${\alpha}$
$\alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx - 1 \right) < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2}$
$< \alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx + 1 \right)$
$\sqrt{\pi} - \alpha < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} < \sqrt{\pi} + \alpha$
If $\alpha \to 0$, both sides tend to $\sqrt{\pi}^2$, and the result follows. In fact, we may use this to find
\begin{align*}
&\left| \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} - \sqrt{\pi} \right| < \alpha \\
&\left| \left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 n^2} dx \right)^2 - \pi \right| < \alpha(\alpha + 2 \sqrt{\pi}) \approx 2 \alpha \sqrt{\pi}.
\end{align*}
Where the approximation is valid for small $α$ similar to the given value $α=1/10^5$
Best Answer
Following this answer of mine, introduce $\omega=\exp(2\pi i/N)$ and consider $$S=\sum_{k=0}^{N-1}\frac{\exp(2nk\pi i/N)}{1-2z\cos(2k\pi/N)+z^2}=\sum_{k=0}^{N-1}\frac{\omega^{kn}}{(1-z\omega^k)(1-z\omega^{-k})}$$ where $|z|\neq 1$ for simplicity (you're computing $3S/N$ at $z=2$). Now use $$\sum_{p=0}^{N-1}(z\omega^{\pm k})^p=\frac{1-(z\omega^{\pm k})^N}{1-z\omega^{\pm k}}=\frac{1-z^N}{1-z\omega^{\pm k}}$$ to get $$(1-z^N)^2 S=\sum_{k=0}^{N-1}\omega^{kn}\sum_{p=0}^{N-1}(z\omega^k)^p\sum_{q=0}^{N-1}(z\omega^{-k})^q=\sum_{p,q=0}^{N-1}z^{p+q}\sum_{k=0}^{N-1}\omega^{k(n+p-q)}.$$
The inner sum is $N$ if $n+p-q$ is a multiple of $N$ (assuming $\color{blue}{0\leqslant n<N}$, this happens iff $0\leqslant p<N-n$ and $q=p+n$, or $N-n\leqslant p<N-n$ and $q=p+n-N$).
Otherwise, this sum is $0$. And finally $$S=\frac{N}{(1-z^N)^2}\left(\sum_{p=0}^{N-n-1}z^{2p+n}+\sum_{p=N-n}^{N-1}z^{2p+n-N}\right)=\frac{N(z^n+z^{N-n})}{(1-z^2)(1-z^N)}$$ (note that $n=N$ is also acceptable here, so that the result holds for $0\leqslant n\leqslant N$).