I don't know why Mathematica does not get it right. At least, it can compute
FourierTransform[Exp[-Sqrt[x.x]], x, z]
1/(1 + z1^2 + z2^2)^(3/2)
From there we can continue on paper:
Let $\varPhi \colon \mathbb{R}^n \to \mathbb{R}^n$ be a diffeomorpism of class at least $C^1$ and denote by $\mathcal{F}$ the Fourier transform. By the transformation formular for integrals (and by certain denseness and continuous extension arguments) one obtains
$$\begin{aligned}\mathcal{F}(f \circ \varPhi)(\xi) &= C \int_{\mathbb{R}^n} f(\varPhi(x)) \, \exp(- \mathrm{i} \, \langle\xi , x \rangle) \, \mathrm{d} x\\&= C \int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle\xi , \varPhi^{-1}(y) \rangle) \, |\det(D\varPhi(y))|^{-1}\, \mathrm{d} y\end{aligned}$$
with a constant $C$ that depends on some convention; the precise value won't matter in the end.
In case of an affine mapping $\varPhi(x) = L \, (x - \mu) = L\, x - L \, \mu$ with $L^T \, L =A$, we obtain $|\det(D\varPhi(y))| = \sqrt{\det(A)}$. Thus, assuming that $A$ is positive definite, we obtain
$$\begin{aligned}\mathcal{F}(f \circ \varPhi)(\xi) &= C\,\det(A)^{-\frac{1}{2}} \int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle\xi , L^{-1}\,y + \mu \rangle) \, \mathrm{d} y\\&= C \, \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle)\int_{\mathbb{R}^n} f(y) \, \exp(- \mathrm{i} \, \langle L^{-T}\xi , y\rangle) \, \mathrm{d} y\\&= \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle) \,\mathcal{F}(f)(L^{-T}\xi).\end{aligned}.$$
With $f(x) = \exp(-\sqrt{\langle x, x\rangle})$ and $$\mathcal{F}(f)(\xi) = (1+\langle \xi, \xi\rangle)^{-\frac{3}{2}},$$
we should obtain $$\mathcal{F}(f \circ \varPhi)(\xi) = \det(A)^{-\frac{1}{2}} \, \exp(- \mathrm{i} \, \langle\xi , \mu \rangle) \, (1+\langle \xi, A^{-1}\xi\rangle)^{-\frac{3}{2}}.$$
But of course, I could have made a mistake. Please check thoroughly.