I started this before @ll-3-14 posted a much more deft answer, but I got far enough that I decided not to discard mine.
Tempered distribution boilerplate
Let
$\varphi\in\mathcal{S}(\mathbb{R}^3)$, and let
$\mathsf{T}$ be the tempered distribution defined by
\begin{equation}
\mathsf{T}[\varphi] = \int_{\mathbb{R}^3}\frac{\varphi(x)}{\left\Arrowvert x\right\Arrowvert^2}dV.
\end{equation}
As a tempered distribution,
$\mathsf{T}$ has a Fourier transform
$\widehat{\mathsf{T}}$ defined by
\begin{equation}
\widehat{\mathsf{T}}[\varphi] = \mathsf{T}[\widehat{\varphi}] = \int_{\widehat{\mathbb{R}}^3}\frac{\widehat{\varphi}(k)}{\left\Arrowvert k\right\Arrowvert^2}dV
\end{equation}
for each
$\varphi\in\mathcal{S}(\mathbb{R}^3)$. I write that
$\widehat{\varphi}\in\mathcal{S}(\widehat{\mathbb{R}}^3)$ to indicate that the Fourier transform
$\widehat{\varphi}$ is a function on the "conjugate space"
$\widehat{\mathbb{R}}^3$.
Beginning the transformation
The goal is to express this as an integral (or limit of integrals) involving
$\varphi$ instead of
$\widehat{\varphi}$.
\begin{equation}
\begin{split}
\int_{\widehat{\mathbb{R}}^3}\frac{\widehat{\varphi}(k)}{\left\Arrowvert k\right\Arrowvert^2}dV(k)
=
\int_{\widehat{\mathbb{R}}^3}\frac{1}{\left\Arrowvert k\right\Arrowvert^2}\left[\int_{\mathbb{R}^3}\varphi(x)e^{-ik\cdot x}dV(x)\right]dV(k),
\end{split}
\end{equation}
where $dV(k)$ is the "differential volume" at $k\in\widehat{\mathbb{R}}^3$ and $dV(x)$ is the "differential volume" at $x\in\mathbb{R}^3$.
\begin{equation}
\begin{split}
&~\lim_{R\to\infty}\int_{\{k:\left\Arrowvert k\right\Arrowvert\leq R\}}\frac{1}{\left\Arrowvert k\right\Arrowvert^2}\left[\int_{\mathbb{R}^3}\varphi(x)e^{-ik\cdot x}dV(x)\right]dV(k)\\
&=~
\lim_{R\to\infty}\int_{\mathbb{R}^3}\varphi(x)\left[\int_{\{k:\left\Arrowvert k\right\Arrowvert\leq R\}}\frac{e^{-i x\cdot k}}{\left\Arrowvert k\right\Arrowvert^2}dV(k)\right]dV(x)
\end{split}
\end{equation}
Let $r = \left\Arrowvert k\right\Arrowvert$. In 3-dimensional spherical co-ordinates,
\begin{equation}
dV(k) = r^2\sin\theta d\theta~dr~d\phi,
\end{equation}
where $\theta$ is the polar angle (varying from 0 to $\pi$ radians) and $\phi$ is the azimuthal angle (varying from 0 to $2\pi$ radians).
The secret
The only angular dependence of the integrand is the dependence on the angle between
$x$ and
$k$.
Define the co-ordinate system so that $x$ points along the $z$-axis, so that
\begin{equation}
x\cdot k = \left\Arrowvert x\right\Arrowvert\left\Arrowvert k\right\Arrowvert\cos\theta = \left\Arrowvert x\right\Arrowvert r\cos\theta,
\end{equation}
where
$\theta$ is the polar angle. The integral of interest is now
\begin{equation}
\int_{0}^{2\pi}\left[\int_{0}^{R}\left[\int_{0}^{\pi}\frac{e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}}{r^2}\sin\theta d\theta\right]r^2dr\right]d\phi
=
2\pi\int_{0}^{R}\left[\int_{0}^{\pi}e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}\sin\theta d\theta\right]dr.
\end{equation}
Calculus!
Let
$u = \left\Arrowvert x\right\Arrowvert r\cos\theta$, so that
\begin{equation}
du = -\left\Arrowvert x\right\Arrowvert r\sin\theta~d\theta,
\end{equation}
and
\begin{equation}
\begin{split}
\int_{0}^{\pi}e^{-i\left\Arrowvert x\right\Arrowvert r\cos\theta}\sin\theta d\theta
&=~
\int_{u=\left\Arrowvert x\right\Arrowvert r}^{u=-\left\Arrowvert x\right\Arrowvert r} e^{-iu}\frac{(-1)du}{\left\Arrowvert x\right\Arrowvert r}\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\int_{u=-\left\Arrowvert x\right\Arrowvert r}^{u=\left\Arrowvert x\right\Arrowvert r}e^{-iu}du\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\left(\left.\frac{e^{-iu}}{-i}\right|_{u=-\left\Arrowvert x\right\Arrowvert r}^{u=\left\Arrowvert x\right\Arrowvert r}\right)\\
&=~
\frac{1}{\left\Arrowvert x\right\Arrowvert r}\left(\frac{-2i\sin(\left\Arrowvert x\right\Arrowvert r)}{-i}\right)\\
&=~ \frac{2\sin(\left\Arrowvert x\right\Arrowvert r)}{\left\Arrowvert x\right\Arrowvert r}.
\end{split}
\end{equation}
A sincing feeling
The integral of interest is now
\begin{equation}
4\pi\int_{0}^{R}\textrm{sinc}(\left\Arrowvert x\right\Arrowvert r)dr,
\end{equation}
where we are using the
un-normalized sinc function. We perform another change of variable:
\begin{equation}
v = \left\Arrowvert x\right\Arrowvert r,~~~\textrm{so that}~~~dr = \frac{dv}{\left\Arrowvert x\right\Arrowvert}.
\end{equation}
The integral is
\begin{equation}
\frac{4\pi}{\left\Arrowvert x\right\Arrowvert}\int_{0}^{R/\left\Arrowvert x\right\Arrowvert}\textrm{sinc}(v)dv.
\end{equation}
There was no other specific
$R$-dependence, and we are taking limits anyway with integrals "against" Schwartz functions, so we can replace
$R/\left\Arrowvert x\right\Arrowvert$ with, say,
$\rho$. We have
\begin{equation}
\widehat{\mathsf{T}}[\varphi]
=
\lim_{\rho\to\infty}\int_{\mathbb{R}^3}\varphi(x)\frac{4\pi}{\left\Arrowvert x\right\Arrowvert}\left[\int_{0}^{\rho}\textrm{sinc}(v)dv\right]dV(x).
\end{equation}
