How to proceed with evaluating $\int_{-\infty}^{\infty}\frac{\sin^n(x)}{x^n}\mathrm{d}x$? 
Exercise. Evaluate the improper trigonometric integral $$\int_{-\infty}^{\infty}\frac{\sin^n(x)}{x^n}\mathrm{d}x, \ \ \ n\in\Bbb{N}^+,$$
  using the complex epsilon method.

Let us view only odd powers, that is
$$\int_{-\infty}^{\infty}\frac{\sin^{2n+1}(x)}{x^{2n+1}}\mathrm{d}x.$$
I have shown that
$$\sin^{2n+1}(x) = \frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sin[(2n+1-2k)x]$$
which leaves us with evaluating
$$\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\Im\overbrace{\left\{\frac{1}{z^{2n+1}+\varepsilon^{2n+1}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\exp[\mathrm{i}(2n+1-2k)z]\right\}}^{f(z)}\mathrm{d}z.$$
We have a total of $n+1$ simple poles of interest, the first $n$ lay in the upper half of the complex plane while the $(n+1)$th is on the real axis. 
$$z_m = \varepsilon\exp\left[\mathrm{i}\frac{(2m+1)\pi}{2n+1}\right], \ \ \ m\in\{0,\ldots,n-1\}.$$ 
Thus
$$\int_{-\infty}^{\infty}\frac{\sin^{2n+1}(x)}{x^{2n+1}}\mathrm{d}x = \lim\limits_{\varepsilon\to0^+}\Im\left[2\pi \mathrm{i}\sum_{m=0}^{n-1}\operatorname{\underset{z=z_m}{Res}}f(z) + \pi\mathrm{i}\operatorname{\underset{z=z_n}{Res}}f(z)+  \lim\limits_{R\to\infty}\int_{C_R}f(z)\right].\tag1$$
From $(1)$ (since simple poles only)
$$\operatorname{\underset{z=z_m}{Res}}f(z)=\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\exp[\mathrm{i}(2n+1-2k)z_m]$$ or, equivalently,
$$\operatorname{\underset{z=z_m}{Res}}f(z)=\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{\infty}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}.$$
The terms of the infinite sum above indice $2n$ become irrelevant as $\varepsilon \to 0^+$. Therefore,
$$\operatorname{\underset{z=z_m}{Res}}f(z)\equiv\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{2n}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}.\tag2$$



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*TL; DR: What is next? How can one simplify the expression


$$\sum_{m=0}^{n-1}\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{2n}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}?\tag3$$
EDIT: I have now accepted an answer that gets around this sum but yet still uses a similar method. However, note that simplifying $(3)$ as a sum remains an open problem for me, so an answer along that line would be entitled to a bounty (+50).
 A: Once you get the Fourier sine series of $\sin(x)^n$, you just need to apply integration by parts multiple times, in order to convert the original integral in a combination of integrals of the form
$$ \int_{-\infty}^{+\infty}\frac{\sin(mx)}{x}\,dx = \pi.$$
A: It is easily seen that for some $\alpha$ $$\left(\frac{\sin x}{x - i\alpha}\right)^{2n+1}=\frac{(-1)^{n}}{i 2^{2n+1}}\sum_{k=0}^{2 n +1}(-1)^k {{2 n +1}\choose {k}} \frac{\exp [i(2\overline{n-k} + 1)x]}{(x - i \alpha)^{2n+1}}.$$ The required integral is the limit $\left[\int_{-\infty}^{\infty}\left(\frac{\sin x}{x - i\alpha}\right)^{2n+1}dx\right]_{\alpha\rightarrow 0^+}$. For the contour closing in the upper half plane, only terms upto $k = n$ in the above sum contribute to the value of the integral. One may rewrite $\exp [i(2\overline{n-k} + 1)x]$ as $\exp [i(2\overline{n-k} + 1)(x - i \alpha)]\exp[-(2\overline{n-k} + 1)\alpha]$ and the residue of each sumand such as $$\frac{\exp [i(2\overline{n-k} + 1)(x - i \alpha)]}{(x - i \alpha)^{2n+1}}$$ is clearly $\frac{i^{2n}(2\overline{n -k}+1)^{2n}}{(2n)!}$. After taking the relevant limit the integral reduces to the following sum $$\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^{2n+1}dx = \frac{\pi}{(2n)!}\sum_{k=0}^{n}(-1)^k{{2n + 1}\choose {k}} \left(n-k+1/2\right)^{2n}$$ which seems simpler than what you obtained.
EDIT: As noted by the OP, the result can be immediately generalized for any positive integer $n$ to $$\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^n dx = \frac{\pi}{2^{n-1}(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k{{n}\choose {k}} \left(n-2k\right)^{n-1}.$$
