What is $\sum_{k=1}^\infty \rm{sinc}^8(k)$ using the sine cardinal function?

Using Bernoulli polynomials, one can make a general formula: $$S_n=\sum_{k=1}^{\infty}\frac{\sin^n k}{k^n}=-\frac{\pi^n}{2n!}\sum_{k=0}^{n}(-1)^k\binom{n}{k}B_n\left(\Big\{\frac{n-2k}{2\pi}\Big\}\right),$$ where $\{x\}=x-\lfloor x\rfloor$ denotes fractional part of $x$. Say, continuing the examples, $$S_{10}=-\frac{1}{2}-\frac{1093\pi}{672}+\frac{5883\pi^2}{896}-\frac{2449\pi^3}{288}+\frac{563\pi^4}{96}\\-\frac{1423\pi^5}{576}+\frac{43\pi^6}{64}-\frac{103\pi^7}{864}+\frac{3\pi^8}{224}-\frac{\pi^9}{1152}+\frac{\pi^{10}}{40320}.$$ BTW, $n=7$ is the first with $n>2\pi$, which causes the complication.

Fourier Analytic Approach

The Fourier Transform of $\frac{\sin(x)}x$ is $$ f(x)=\pi\!\left[-\tfrac1{2\pi}\le\xi\le\tfrac1{2\pi}\right]\tag1 $$ This would mean that the Fourier Transform of $\frac{\sin^n(x)}{x^n}$ is $f_n(\xi)=\left(\ast^n\right)\!f(\xi)$, which is the convolution of $n$ copies of $f$.

The Poisson Summation Formula says that $$ \sum_{k\in\mathbb{Z}}\frac{\sin^n(k)}{k^n}=\sum_{k\in\mathbb{Z}}f_n(k)\tag2 $$ The support of $f$ is $\left[-\frac1{2\pi},\frac1{2\pi}\right]$; therefore, the support of $f_n$ is $\left[-\frac{n}{2\pi},\frac{n}{2\pi}\right]$. Furthermore, since $f$ is even, $f_n$ is also. Thus, $$ \sum_{k=1}^\infty\frac{\sin^n(k)}{k^n}=\frac{f_n(0)-1}2+\sum_{k=1}^{\left\lfloor\frac{n}{2\pi}\right\rfloor}f_n(k)\tag3 $$ For $n\le6$, the right side of $(3)$ is $\frac{f_n(0)-1}2$. For $7\le n\le12$, the right side of $(3)$ is $\frac{f_n(0)-1}2+f_n(1)$. For $13\le n\le18$, the right side of $(3)$ is $\frac{f_n(0)-1}2+f_n(1)+f_n(2)$. And so on.

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Contour Integration

We can use contour integration to get $$ \begin{align} f_n(\xi) &=\int_{-\infty}^\infty\frac{\sin^n(x)}{x^n}e^{-2\pi ix\xi}\,\mathrm{d}x\\ &=\int_{-\infty-i}^{\infty-i}\frac{\left(e^{ix}-e^{-ix}\right)^n}{(2ix)^n}e^{-2\pi ix\xi}\,\mathrm{d}x\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{-\infty-i}^{\infty-i}\frac{e^{i(n-2k-2\pi\xi)x}}{(2ix)^n}\,\mathrm{d}x\\ &=\sum_{k=0}^{\lfloor n/2-\pi\xi\rfloor}(-1)^k\binom{n}{k}2\pi\frac{(n-2k-2\pi\xi)^{n-1}}{2^n(n-1)!}\\ &=\frac{\pi}{2^{n-1}(n-1)!}\sum_{k=0}^{\lfloor n/2-\pi\xi\rfloor}(-1)^k\binom{n}{k}(n-2k-2\pi\xi)^{n-1}\tag4 \end{align} $$

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Computation

Applying $(4)$ to $(3)$, we can compute $\sum\limits_{k=1}^\infty\frac{\sin^n(k)}{k^n}$ for any $n$: $$ \begin{array}{l|l} n&\sum\limits_{k=1}^\infty\frac{\sin^n(k)}{k^n}\\\hline 1&\frac{\pi-1}2\\ 2&\frac{\pi-1}2\\ 3&\frac{3\pi-4}8\\ 4&\frac{2\pi-3}6\\ 5&\frac{115\pi-192}{384}\\ 6&\frac{11\pi-20}{40}\\ 7&\frac{5887\pi-11520}{23040}+\frac{\pi(7-2\pi)^6}{46080}\\ 8&\frac{151\pi-315}{630}+\frac{\pi(4-\pi)^7}{5040}\\ 9&\frac{259723\pi-573440}{1146880}+\frac{\pi(9-2\pi)^8}{10321920}-\frac{\pi(7-2\pi)^8}{1146880}\\ 10&\frac{15619\pi-36288}{72576}+\frac{\pi(5-\pi)^9}{362880}-\frac{\pi(4-\pi)^9}{36288}\\ 11&\frac{381773117\pi-928972800}{1857945600}+\frac{\pi(11-2\pi)^{10}}{3715891200}-\frac{11\pi(9-2\pi)^{10}}{3715891200}+\frac{11\pi(7-2\pi)^{10}}{743178240}\\ 12&\frac{655177\pi-1663200}{3326400}+\frac{\pi(6-\pi)^{11}}{39916800}-\frac{\pi(5-\pi)^{11}}{3326400}+\frac{\pi(4-\pi)^{11}}{604800} \end{array} $$