The generalization looks like this $$ \int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2\pi/l,~l\in\mathbb{N}\tag{1} $$$$ \int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2/l,~l\in\mathbb{N}\tag{1} $$ where $n$ need not be an integer. The general theorem is given for example in the paper Surprising sinc sums and integrals. Below I give the general outline of the proof which is based on the well known fact that the following function is band limited (its Fourier transform has limited spectrum) $$g(x)=\binom{n}{x}=\frac{1}{2\pi}\int_{-\pi}^{\pi} (1+e^{i \omega})^n e^{- ix\omega} d\omega$$ One can see that Fourier transform is limited to frequencies $|\omega|<\pi$. Whenever spectrum of a function $f(x)$ is limited to frequencies $|\omega|<2\pi$ one expects that $$ \int_{-\infty}^{\infty} f(x) dx =\sum_{k=-\infty}^\infty f(k). $$
Now the Fourier transform of $g(\alpha x)^l$ has a spectrum limited in the band $|\omega|<\pi\alpha l$. This is easy to see calculating Fourier transform $$ \int_{-\infty}^{\infty}g(\alpha x)^le^{-ikx}dx $$ with the help of $\int_{-\infty}^{\infty}e^{-ikx}dx=2\pi\delta(k)$, where $\delta$ is delta function. The general theorem from the paper cited above now states that $$ \int_{-\infty}^{\infty} g(\alpha x)^l dx =\sum_{k=-\infty}^\infty g(\alpha k)^l,\quad 0<\alpha\le 2\pi/l,~l\in\mathbb{N}. $$$$ \int_{-\infty}^{\infty} g(\alpha x)^l dx =\sum_{k=-\infty}^\infty g(\alpha k)^l,\quad 0<\pi\alpha l\le 2\pi,~l\in\mathbb{N}, $$ Allwhich is equivalent to (1).
All this analysis also explains that when $\alpha=1$ the proposed identity holds for $l=1,2$ but not for larger $l\in\mathbb{N}$.