By definition of Fourier transform (with your normaliztion)
$$
F(\xi):=\mathcal{F}[|x|^{-m}](\xi)=\frac{1}{(2\pi)^{n/2}}\int |x|^{-m }e^{ix\cdot \xi} d^nx
$$
This integral, however, has a lot of problems. So instead let's work with a closely related (convergent) integral:
$$
\begin{aligned}
G(\xi;K,\epsilon)=\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|<\frac{1}{K}} |x|^{-m }(e^{ix\cdot \xi}-1) d^nx\\+
\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|\geq \frac{1}{K}} |x|^{-m }e^{ix\cdot \xi}e^{-\epsilon |x|} d^nx
\end{aligned}
$$
where $K$ is very large and $\epsilon$ is very small. This is designed to avoid the following problems: 1) $F(\xi=0)$ is infinite for $n>m$, because of the behavior near infinity, 2) if $m>n$ the integral $F(\xi)$ is divergent altogether because of the behavior near zero. Note that naively,
$$
\begin{aligned}
F(\xi)&=G(\xi,K,0)+\frac{1}{(2\pi)^{\frac{n}{2}}} \int_{|x|<\frac{1}{K}} |x|^{-m } d^nx \\
&=G(\xi;K,0)+\frac{S_n}{(2\pi)^n}\left.\frac{x^{n-m}}{n-m}\right|_0^{1/K}
\end{aligned}
$$
with $S_n=2\pi^{n/2}/\Gamma(n/2)$ and $\Gamma$ the Gamma function. The second summand is infinity if $m>n$ (in a sense there is "an infinity" embeded in $F(\xi)$ that you cannot get rid of). However, if $n>m$ (note that I'm avoiding $n=m$ case like the plague! All hell breaks loose in that situation! And I personally don't know how to deal wih it), the second summand grows like $\sim K^{-(n-m)}$. In fact, if $n>m$ then one can safely take $K=\infty$. I hope this explains to you why people always only talk about the case $m<n$, because the $m>n$ is badly divergent. Even if $m<n$ the raw Fourier transform is still divergent for $\xi=0$, but that's much more managable! So let's study $G$ now.
Observation 1. Let $R$ be a rotation around the origin. Since $R$ is an isometry, then $G(R\xi;K,\epsilon)=G(\xi;K,\epsilon)$. So $G(\xi;K,\epsilon)=G'(|\xi|;K,\epsilon)$.
Observation 2. Let $\lambda> 0$ be a real number, then
$$
\begin{aligned}
G(\lambda\xi;\lambda K,\lambda\epsilon)&=G'(\lambda|\xi|;\lambda K,\lambda \epsilon)\\
&=
\lambda^{m-n}G'(|\xi|;K,\epsilon)=
\lambda^{m-n}G(\xi;K,\epsilon)
\end{aligned}
$$
Therefore $G'$ is a homogeneous function of degree $m-n$. In case $m<n$, $G'$ is equal to $1/p(|\xi|;K, \epsilon)$ where $p$ is a homogeneous polynomial of degree $n-m$. With assuption $K=0$, and $\xi=0$ you can find that $p(0,0,\epsilon)=\epsilon^{n-m}$, which shows the divergent behavior I was talking about at $\xi=0$. At the same time, since $F(\xi)$ is not identically infinity (i.e. there exist $\xi$ such that $1/F(\xi)\neq 0$), $p(|\xi|,0,0)=|\xi|^{n-m}$.
So up until now, we have shown that away from $\xi=0$ we have $F(\xi)=C(m,n)|\xi|^{m-n}$ for some $C(m,n)\in \mathbb{C}$ if $n>m$. Also $F(\xi)$ is ill-defined if $m>n$. Can we calculate what $C(m,n)$ is? Sure...
Observation 3. Let $f,g$ be two function, $F,G$ their Fourier transform. Note that
$$
(f,g):=\int f(x)g(x)d^nx = \int F(\xi)G(-\xi)d^n\xi
$$
Moreover, $\mathcal{F}(\exp(-|x|^2/2)=\exp(-|\xi|^2/2)$. Now comes the trick
$$
\begin{aligned}
(|x|^{-m},\exp(-|x|^2/2)&=\int |x|^{-m}\exp(-|x|^2/2)d^nx \\
&= C(m,n)\int
|\xi|^{m-n}\exp(-|\xi|^2/2)d^n\xi\\
\Longrightarrow C(m,n)&=\frac{\int |x|^{-m}\exp(-|x|^2/2)d^nx}{
\int |x|^{m-n}\exp(-|x|^2/2)d^nx}=\boxed{\frac{2^{\frac{n-m}{2}}\Gamma(\frac{n-m}{2})}{2^{\frac{m}{2}}\Gamma(\frac{m}{2})}}
\end{aligned}
$$
This completes the discussion.
