Evaluate $\int_{(-\infty,\infty)^n}\frac{\prod_{k=1}^n \sin(a_k x_k)}{\prod_{k=1}^n x_k}\frac{\sin(\sum_{k=1}^n a_k x_k)}{\sum_{k=1}^n a_k x_k}$

Consider a more general integral ($a_k,b_k,c_k>0$ for $1\leqslant k\leqslant n$): \begin{align}I&:=\int_{(-\infty,\infty)^n}\left(\prod_{k=1}^n\frac{e^{-c_k|x_k|}\sin a_k x_k}{x_k}\right)\frac{\sin\sum_{k=1}^{n}b_k x_k}{\sum_{k=1}^{n}b_k x_k}dx_1\ldots dx_n\\ &=\frac12\int_{(-\infty,\infty)^n}\left(\prod_{k=1}^n\frac{e^{-c_k|x_k|}\sin a_k x_k}{x_k}\right)\int_{-1}^1\exp\left(it\sum_{k=1}^n b_k x_k\right)dt\,dx_1\ldots dx_n\\ &=\frac12\int_{-1}^1\left(\prod_{k=1}^n\int_{-\infty}^\infty e^{-c_k|x_k|}\frac{\sin a_k x_k\cos tb_k x_k}{x_k}\,dx_k\right)dt\\ &=\frac12\int_{-1}^1\prod_{k=1}^n\left(\arctan\frac{a_k+b_k t}{c_k}+\arctan\frac{a_k-b_k t}{c_k}\right)dt. \end{align} The given integral is obtained at $b_k=a_k$ and $c_k\to 0$ (which is allowed under the integral sign, since the convergence is absolute) and is equal to $\color{blue}{\pi^n}$. For arbitrary $b_k$, the answer is $\pi^n\min\big\{1,\min\limits_{1\leqslant k\leqslant n}(a_k/b_k)\big\}$, since the limit of the integrand is $0$ if $b_k|t|>a_k$ for any $k$.

Recall Parseval Identity ($F(k)=\int_R e^{i k t}f(x)$):

$$ \mathcal{I}=\int_R f(x)g(x)=\frac{1}{2\pi}\int_RF(k)\bar{G}(k) $$

take $f(x)=\text{sinc}(x+l), g(x)=\text{sinc}(x)$. The Fourier transform of $f(x)$ is standard

$$ F(k)=\pi e^{-ilk}\chi_{[-1,1]}(k)$$

Consequently $$ \mathcal{I}=\frac{\pi}2\int_{-1}^1 e^{-ilk}=\pi \text{sinc}(l) \quad (\star) $$

Now setting $x=x_1, l=\sum_{n\geq i>1}x_i$ in your integral gives

$$ I_n=\int_{R^{n-1}}dx^{n-1}\prod_{n\geq i>1}(\text{sinc}(x_i))\int_Rdx\,\text{sinc}(x+l)\text{sinc}(x) $$

or (using $(\star)$ and restorting l)

$$ I_n= \pi \int_{R^{n-1}}dx^{n-1}\prod_{n\geq i>1}(\text{sinc}(x_i)) \times \mathcal{I} \\=\pi\int_{R^{n-1}}dx^{n-1}\text{sinc}(\sum_{n\geq i>1}x_i)\prod_{n\geq i>1}(\text{sinc}(x_i)) = \\ \pi I_{n-1} $$

with the starting point $I_1=\pi$ (which also follows from Parseval) you can now solve the (trivial) recurrence:

$$ I_n=\pi^n $$