$ \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x = \frac{\pi}{2^{n+1} \cdot (n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (2k-n)^{n-1} \operatorname{sign}(2k-n) $
where $\operatorname{sign}(x) = \cases{ 1 & $x > 0$ \\ 0 & $x = 0$\\ -1 & $x < 0$}$.
As to the (probabilistic) proof, notice that $\frac{\sin(t)}{t}$ is the characteristic function of a uniform random variable on $(-1,1)$. The sum of $n$ independent identically distributed such uniform random variables is known as Irwin-Hall random variable $Y_n$, and the integral in question is a multiple of its PDF evaluated at $x=0$: $ \phi_{Y_n}(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty \frac{\sin^n(t)}{t^n} \mathrm{e}^{-i t x} \mathrm{d} t = \frac{1}{\pi} \int_{0}^\infty \frac{\sin^n(t)}{t^n} \cos(t x) \mathrm{d} t $ The closed form for the PDF is given on the wikipedia with the reference.
As to more explicit derivation. We first integrate by parts,
$n-1$ times, then use binomial theorem for
$\sin^n(x)$:
$ \begin{eqnarray} \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \int_0^\infty \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \sin^n(x) \right) \frac{1}{(n-1)!}\frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \mathrm{e}^{i (2k-n)x} \right) \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left(i (2k-n)\right)^{n-1} \mathrm{e}^{i (2k-n)x} \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \sin((2k-n)x) \frac{\mathrm{d} x}{x} \end{eqnarray} $ In the last line,
$\mathrm{e}^{i (2k-n) x}$ was expanded use Euler's formula, and since the sum is real, only real summands are retained. Then, integrating term-wise nails it:
$ \begin{eqnarray} \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \int_0^\infty \sin((2k-n)x) \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \frac{\pi}{2} \operatorname{sign}(2k-n) \end{eqnarray} $