Results like this hold for small values of $j$, but are eventually false. Put $f(x)$ to be the indicator function of the interval $[-1/2,1/2]$, and put $f_j(x)$ to be the convolution of $f$ with itself $j$ times. Then the Fourier transform of $f_j$ is
$$
{\hat f_j}(\xi) = \int_{-\infty}^{\infty} f_j(x) e^{-2\pi i x\xi} dx =
\Big( \int_{-1/2}^{1/2} e^{-2\pi i x\xi } dx \Big)^j = \Big(\frac{\sin (\pi \xi)}{\pi \xi} \Big)^j = (\text{sinc}(\pi \xi))^j.
$$

So the sum in the question is
$$
\sum_{n=1}^{\infty} {\hat f_j}(n \alpha_j), \tag{1}
$$
with
$$
\alpha_j = \frac{1}{\pi^2} \int_{{\Bbb R}} (\text{sinc}(x))^j dx.
$$

Since $\text{sinc}$ is even, and with the usual convention that its value at $0$ is $1$, the quantity in (1) is
$$
-\frac 12 + \frac 12 \sum_{n\in {\Bbb Z}} {\hat f}_j(n\alpha_j)= -\frac 12 + \frac{1}{2\alpha_j} \sum_{k} f_j(k/\alpha_j), \tag{2}
$$
by the Poisson summation formula.

The contribution of the term $k=0$ above is
$$
\frac{1}{2\alpha_j} f_j(0) = \frac{1}{2\alpha_j} \int_{-\infty}^{\infty} (\text{sinc}(\pi x))^j dx = \frac{\pi}{2}.
$$
This explains the purported identity: the function $f_j$ is supported in $[-j/2,j/2]$ and for small $j$ one has
$$
\frac{1}{\alpha_j} \ge \frac{j}{2},
$$
so that all terms with $k\neq 0$ in (2) vanish. But it is easy to see that $\alpha_j$ decreases like constant$/\sqrt{j}$, so eventually non zero values of $k$ will contribute, and there will be no exact identity.