Hm, that looks tricky. Looking in the integral book Gradshteyn and Ryzhik, I found two useful integrals. The first is
$$\int_0^\infty dx~\exp(-x^\mu) = \frac{1}{\mu}\Gamma\left(\frac{1}{\mu}\right),$$
which holds for ##\mbox{Re}(\mu) > 0## and can be used to find the normalization constant of your distribution. This is integral 3.326-1 in the sixth edition.
There does not appear to be an integral for ##\exp(-x^\mu+ix)##, so it's possible there may not be a closed form for the Fourier transform. However, you could expand the imaginary exponential in a power series and perform the integral term-by-term to get a power series representation of the Fourier transform. In this case, the following integral (3.326-2) is useful:
$$\int_0^\infty dx~x^m \exp(-\beta x^n) = \frac{\Gamma(\gamma)}{n\beta^\gamma},$$
where ##\gamma = (m+1)/n##.