$$ I_n = 2\int_{0}^{+\infty}\frac{\sin(\pi x)}{\prod_{k=-n}^{n}(x-k)}\,dx $$
By the residue theorem we have the following partial fraction decomposition:
$$ \frac{1}{\prod_{k=-n}^{n}(x-k)}=\sum_{k=-n}^{n}\frac{(-1)^k}{(x-k)}\cdot\frac{\binom{2n}{n+k}}{(2n)!} \tag{0}$$
Recalling $\int_{-\infty}^{+\infty}\frac{\sin(\pi x)}{x\pm k}\,dx=\pi(-1)^k$, the identity $(0)$ allows to state that
$$I_n=\color{red}{\frac{\pi(-1)^n 4^n}{(2n)!}}.\tag{1}$$
Addendum 1: in terms of the (inverse) Laplace transform, $(0)$ is equivalent to:
$$\mathcal{L}^{-1}\left( \frac{1}{\prod_{k=-n}^{n}(x-k)}\right)(s)=\sum_{k=-n}^{n}\frac{(-1)^k}{(n-k)!(n+k)!}e^{ks}=\frac{4^n}{(2n)!}\left(\sinh \tfrac{s}{2}\right)^{2n}. \tag{2}$$
By the shift property of $\mathcal{L}^{-1}$, we may state the same as
$$\mathcal{L}^{-1}\left( \frac{1}{\prod_{k=0}^{2n}(x+k)}\right)(s)=\frac{4^n}{(2n)!}\left(\sinh \tfrac{s}{2}\right)^{2n} e^{-ns} = \frac{4^n}{(2n)!}\left(1-e^{-s}\right)^{2n}. \tag{3}$$
Addendum 2: by the Weierstrass product for the sine function,
$$ \frac{\sin(\pi x)}{\prod_{k=-n}^{n}(x-k)}\approx \frac{\pi(-1)^n}{n!^2}\exp\left[-(\zeta(2)-H_n^{(2)})x^2\right] $$
hence this exercise also leads to an interesting approximation:
$$ \frac{1}{4^n}\binom{2n}{n}\geq \sqrt{\frac{\zeta(2)-H_n^{(2)}}{\pi}}$$
which is a nice counter-part to $\frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi n}}$.
Addendum 3: the identity also follows from the continuous analogue of the binomial theorem
$$ \int_{-\infty}^{+\infty}\binom{n}{x}\,dx = 2^n $$
proved on MO.
