Using the Fourier transform of $e^{- \alpha \lvert t \rvert}$, we find
\begin{align}
F(\omega) &= \frac{2 \alpha}{\alpha^{2} + \omega^{2}}
\end{align}
Hence,
\begin{align}
e^{- \alpha \lvert t \rvert} &= \frac{1}{2 \pi} \int_{\mathbb{R}} \frac{2 \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\
\implies \pi e^{- \alpha \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ \alpha}{\alpha^{2} + \omega^{2}} e^{i \omega t} d \omega \\
\end{align}
Now, set $\alpha = 1$ and map $t \mapsto -t$
\begin{align}
\implies \pi e^{- \lvert -t \rvert} &= \pi e^{- \lvert t \rvert} \\
&= \int_{\mathbb{R}} \frac{ 1}{1 + \omega^{2}} e^{- i \omega t} d \omega \\
\end{align}
Now map $\omega \mapsto \omega + 1$
\begin{align}
\implies \pi e^{- \lvert t \rvert} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i (\omega + 1) t} d \omega \\
&= e^{-i t} \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega \\
\implies \pi e^{- \lvert t \rvert + i t} &= \int_{\mathbb{R}} \frac{ 1}{1 + (\omega + 1)^{2}} e^{- i \omega t} d \omega
\end{align}
hence, by the Fourier inversion formula
\begin{align}
F(\omega) &= \int_{\mathbb{R}} \frac{ 1}{1 + (t + 1)^{2}} e^{- i \omega t} dt \\
&= \pi e^{i \omega - \lvert \omega \rvert}
\end{align}
Use the following analogue of the Cauchy estimates for harmonic functions: if $u:\mathbb{R}^n\to\mathbb{R}$ is harmonic, then for any multi-index $\alpha$ and $r>0$,
\begin{equation}
\vert D^{\alpha}(0)\vert\leq \frac{C_{\alpha}\sup_{\vert x\vert=r} \vert u(x)\vert}{r^{\vert \alpha\vert}},
\end{equation}
where $C_{\alpha}$ is some constant depending only on $\alpha$, and $\vert \alpha\vert$ is the order of $\alpha$. For any multi-index $\alpha$ with $\vert \alpha\vert\geq m+1$, your assumption shows that
\begin{equation}
\vert D^{\alpha}(0)\vert\leq \frac{C'_{\alpha}r^m}{r^{m+1}}\to 0,
\end{equation}
where $C'_{\alpha}$ is some other constant depending only on $\alpha$ as $r\to \infty$, so every partial derivative at $0$ of order at least $m+1$ is zero. For any other $x\in \mathbb{R}^n$, simply shift the function appropriately and use the same argument to show all derivatives of order $\geq m+1$ are zero at $x$.
The reason why the Cauchy estimates hold is by writing $u$ using the Poisson kernel and differentiating under the integral: see page 33 of these notes for a full proof.
Best Answer
Of course we treat this as a tempered distribution, since it's not in $L^1$.
As your notation already maybe suggests, writing the function as ${1\over |x|^s}P\Big({x\over |x|}\Big)$ usefully separates the radial and spherical coordinates.
Fourier transform behaves well with respect to homogeneity, sending a tempered distribution of degree $-s$ (thinking of $1/|x|^s$ to normalize) to a tempered distribution of degree $-(\hbox{dim}-s)$. Fourier transform commutes with rotations. We do somehow know that the space of harmonic degree $d$ polynomials (with or without dividing by $|x|^d$) is an irreducible representation of the rotation group, so (Schur's lemma) the only automorphisms of that space, commuting with rotations, are scalars.
Thus, up to some constant $C=C(s,n,d)$, the Fourier transform of the (tempered distribution, possibly requiring meromorphic continuation for some values of $s\in\mathbb C$), is $$ C\cdot {1\over |x|^{(n+1)-s}}\cdot P\bigg({x\over |x|}\bigg) \;=\; C \cdot {1\over |x|^{(n+1)+d-s}} \cdot P(x) $$ At $s=n$, this is $$ C\cdot {1\over |x|^{1+d}}\cdot P(x) $$ The constant(s) can be determined by evaluation at aptly chosen Schwartz functions, such as $P(x)e^{-\pi |x|^2}$.
EDIT: in fact, since Fourier transform is an isomorphism on spaces of harmonic functions, and these are irreducible, it suffices to take $P(x_1,\ldots,x_n)=(x_1+ix_2)^k$, which reduces the computation of the constant to a quite tractable elementary issue.