Upon request of user409452, here is an extended version of my comment to Alexandre Eremenko's answer.

First of all, for $a > 0$ we evaluate
$$ \phi(a) = \int_0^\infty e^{-a^2 t/4} t^{-3/2} e^{-1/t} dt . $$
By the dominated convergence theorem,
$$ \phi'(a) = -\frac{a}{2} \int_0^\infty e^{-a^2 t/4} t^{-1/2} e^{-1/t} dt . $$
On the other hand, substituting $t = 4/(a^2 s)$ we find that
$$ \phi(a) = \frac{a}{2} \int_0^\infty e^{-1/s} s^{-1/2} e^{-a^2 s / 4} ds . $$
It follows that
$$ \phi'(a) = -\phi(a) , $$
and hence
$$ \phi(a) = e^{-a} \phi(0) . $$
By a substitution $t = 1/s$,
$$ \phi(0) = \int_0^\infty t^{-3/2} e^{-1/t} dt = \int_0^\infty s^{-1/2} e^{-s} ds = \Gamma(\tfrac12) = \sqrt{\pi} , $$
and so we finally have
$$ \phi(a) = \sqrt{\pi} e^{-a} . $$

Now we write
$$ f(\xi) = e^{-|\xi|} = \pi^{-1/2} \phi(|\xi|^2) = \frac{1}{\sqrt{\pi}} \int_0^\infty e^{-|\xi|^2 t / 4} t^{-3/2} e^{-1/t} dt . $$
Evaluating the inverse Fourier transform of both sides and using Fubini's theorem, we find that
$$ \check{f}(x) = \frac{1}{\sqrt{\pi}} \int_0^\infty \frac{1}{(\pi t)^{n/2}} e^{-|x|^2 / t} t^{-3/2} e^{-1/t} dt = \frac{1}{\pi^{(n + 1)/2}} \int_0^\infty t^{-(3 + n) / 2} e^{-(1 + |x|^2) / t} dt . $$
Once again we substitute $t = 1 / s$:
$$ \check{f}(x) = \frac{1}{\pi^{(n + 1)/2}} \int_0^\infty s^{(n - 1)/2} e^{-(1 + |x|^2) s} ds = \frac{1}{\pi^{(n + 1)/2} (1 + |x|^2)^{(n + 1)/2}} \, \Gamma(\tfrac{n + 1}{2}) . $$
Finally,
$$ \hat{f}(x) = (2 \pi)^n \check{f}(x) = 2^n \Gamma(\tfrac{n + 1}{2}) \pi^{(n - 1)/2} \, \frac{1}{(1 + |x|^2)^{(n + 1)/2}} \, . $$