Although this can be proven using the fact that the Fourier transform is homogeneous, I think it's quite interesting to see a more constructive proof. Also, let me write this in parts, so that the details do not obscure the overall proof. Also, I will provide 2 different proofs, corresponding to different regularizations, but ultimately give the same answer. The first method is easier intuitively but harder to rigorously regularize, while the 2nd method is easier to rigorously prove, but requires a special trick.
Method 1 (Contour integrals)
Non-rigorous sketch. Notice that by rotating $k$, we see that the inverse Fourier transform is given by
\begin{equation}
\int \frac{1}{k^2} e^{ikx} \frac{dk}{(2\pi)^d}=\int \frac{1}{k^2}e^{ik_1 r} \frac{dk}{(2\pi)^d}, \quad r=|x|
\end{equation}
Let us write $k=(p,q)$ where $p=k_1$ and $q=(k_2,...,k_d) \in \mathbb{R}^{d-1}$. Then we have
$$
\int \frac{1}{p^2+q^2} e^{ipr} \frac{dpdq}{(2\pi)^d}= \int \frac{dq}{(2\pi)^{d-1}}\int \frac{1}{p^2+q^2} e^{ipr} \frac{dp}{2\pi}
$$
(This is a non-rigorous application of the Fubini theorem since the integrand is not absolutely integrable). The inner integral can be computed using contour integrals so that for all $q\ne 0$, we have
$$
\int \frac{1}{p^2+q^2} e^{ipr} \frac{dp}{2\pi}= \frac{1}{2|q|} e^{-|q|r}
$$
Plug this back in and using spherical coordinates, we have
\begin{align}
\int_0^\infty \frac{\rho^{d-2} d\rho}{(2\pi)^{d-1}} \frac{1}{2\rho } e^{-\rho r}\cdot S_{d-2} &= \frac{1}{r^{d-2}}\cdot\frac{1}{2(2\pi)^{d-1}} S_{d-2}\int _0^\infty z^{d-3}e^{-z}dz \\
&= \frac{1}{r^{d-2}}\cdot\frac{1}{2(2\pi)^{d-1}} S_{d-2} \cdot(d-3)! \\
\end{align}
where $S_{d-2}$ is the surface area of the unit sphere in $d-1$-dimensions. Using the Legendre duplication formula, you can check that this is exactly that given in Wiki.
Rigorous Proof. Let me only care about $d>2$ so that $1/k^2$ is locally integrable in $\mathbb{R}^d$. Then the inverse Fourier transform of $1/k^2$ as a tempered distribution is given by
$$
\int \frac{1}{k^2} \check{\varphi}(k) \frac{dk}{(2\pi)^d} = \int \frac{1}{k^2} \frac{dk}{(2\pi)^d}\int\varphi(x)e^{-ikx} \frac{dk}{(2\pi)^d}
$$
where $\check{\varphi}(x)$ is the inverse Fourier transform of $\varphi\in S$ in the Schwartz space.
Now the hard part is choosing the right regularization so that we can apply Fubini's theorem. Let $n\in \mathbb{N}$ be such that $n >(d-2)/2$. Then we see that
$$
\frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n
$$
is integrable in $\mathbb{R}^d$ and $\to 1/k^2$ in $S'$ as $\Lambda \to \infty$. Since the Fourier transform is continuous on $S'$, we see that we can just compute the Fourier transform of the regularization and take $\Lambda \to \infty$. Now consider
$$
\int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ikx} \frac{dk}{(2\pi)^d} = \int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ik_1 r} \frac{dk}{(2\pi)^d}, \quad r=|x|
$$
where we rotated $k$ so that we can replace $kx$ with $k_1r$. Now for simplicity, let $k=(p,q)$ where $p=k_1$ and $q=(k_2,...,k_d)$. Then we see that
$$
\int \frac{1}{k^2} \left(\frac{\Lambda^2}{k^2+\Lambda^2}\right)^n e^{ikx} \frac{dk}{(2\pi)^d} = \int \frac{dq}{(2\pi)^{d-1}} \int \frac{dp}{2\pi} \frac{1}{p^2+q^2} \left( \frac{\Lambda^2}{p^2+q^2+\Lambda^2}\right)^n e^{ipr}, \quad r=|x|
$$
Using contour integrals, we see that
$$
\int \frac{dp}{2\pi} \frac{1}{p^2+q^2} \left( \frac{\Lambda^2}{p^2+q^2+\Lambda^2}\right)^n e^{ipr} = \frac{C_n \Lambda^{2n-2}}{(q^2+\Lambda^2)^{(2n-1)/2}} e^{-r\sqrt{q^2+\Lambda^2}} +\frac{1}{2|q|} e^{-|q|r}
$$
Now the first term $\to 0$ as $\Lambda \to\infty$ and thus by dominated convergence, the statement follows as in the non-rigorous sketch.
Method 2 (Gaussian decay)
Let me only care about $d>2$ so that $1/k^2$ is locally integrable. Notice that $e^{-\epsilon k^2}/k^2 \to 1/k^2$ as tempered distributions. Hence, we only need to compute the Fourier transform of $e^{-\epsilon k^2}/k^2$ and take $\epsilon \to 0$, i.e.,
\begin{align}
\int \frac{e^{ikx}}{k^2} e^{-\epsilon k^2} \frac{dk}{(2\pi)^d} &= \int \frac{dk}{(2\pi)^d} e^{ikx} \int_\epsilon^\infty e^{-tk^2} dt \\
&= \int_\epsilon^\infty dt \int \frac{dk}{(2\pi)^d} e^{-tk^2+ikx}\\
&= \frac{1}{(4\pi)^{d/2}}\int_\epsilon^\infty t^{-d/2} e^{-x^2/4t} dt \\
&= \frac{1}{|x|^{d-2}} \frac{1}{2^2 \pi^{d/2}} \int_0^{x^2/4\epsilon} e^{-y} y^{d/2-2} dy \\
&= \frac{1}{|x|^{d-2}} \frac{1}{2^2 \pi^{d/2}} \Gamma \left(\frac{d}{2}-1 \right) , \quad \epsilon \to 0
\end{align}
