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In short, in $\mathbb{R}^d$, if I define the Fourier transform as $\mathcal{F}(f)(x) = \int_{\mathbb{R}^d} f(y) \,e^{-2iπxy}\,\mathrm{d}y$ the result is
$$
\boxed{\mathcal{F}\left(\frac{1}{\omega_d\,|x|^d}\right) = \frac{\psi(d/2)-\gamma}{2} - \ln(|πx|)}
$$
where $\omega_d = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere (so $\omega_3 = 4\pi$), $\gamma$ is Euler–Mascheroni constant and $\psi$ is the digamma function. Since $\psi(3/2) = 2-\gamma-\ln(4)$, we deduce that in dimension $3$
$$
\mathcal{F}\left(\frac{1}{4\pi\,|x|^3}\right) = 1-\gamma -\ln(|2πx|)
$$
which with your Fourier transform convention gives
$$
\boxed{\tilde{\mathcal{F}}\left(\frac{1}{|x|^3}\right) = \frac{1-\gamma -\ln(|x|)}{2\pi^2}.}
$$

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Now the details. So first, what is the meaning of $\frac{1}{|x|^d}$? One can define the following distribution
$$
v_d := \mathrm{pf}\left(\frac{1}{|x|^d}\right) := \mathrm{div}\left(\frac{x\ln(|x|)}{|x|^d}\right)
$$
where the derivative is taken in the sense of distributions. One easily verifies that the distribution $v_d$ coincides on $\Bbb R^d\setminus\{0\}$ with the function
$$
v_d(x) = \frac{1}{|x|^d} \text{ for any } x≠ 0.
$$
One also notices that it is a tempered distribution as the derivative of a function in $L^1 + L^\infty$.

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Let $u_d = \frac{x\ln(|x|)}{|x|^d}$. Multiplying by a test function $\varphi\in C^\infty_c$, one gets for any $\lambda>0$
$$
\begin{align*}
\langle v_d,\varphi\rangle &= -\int_{\mathbb{R}^d} u_d\cdot\nabla\varphi
\newline
&= -\int_{|x|\leq\lambda} u_d\cdot\nabla(\varphi(x)-\varphi(0)) - \int_{|x|>\lambda} u_d\cdot\nabla \varphi
\end{align*}
$$
which by integration by parts yields
$$
\begin{align}\tag{1}
\langle v_d,\varphi\rangle &= \int_{|x|\leq\lambda} \frac{\varphi(x)-\varphi(0)}{|x|^d}\,\mathrm{d}x
\newline
&\quad+ \int_{|x|>\lambda} \frac{\varphi(x)}{|x|^d} \,\mathrm{d}x + \omega_d \ln(\lambda) \,\varphi(0).
\end{align}
$$
One can take $\lambda = 1$ to get
$$
\begin{align*}
\langle v_d,\varphi\rangle &= \int_{|x|\leq 1} \frac{\varphi(x)-\varphi(0)}{|x|^d}\,\mathrm{d}x + \int_{|x|> 1} \frac{\varphi(x)}{|x|^d} \,\mathrm{d}x.
\end{align*}
$$
But with the formula (1) with $\lambda\neq 1$, one can also compute easily the distribution $v_d(\lambda x)$ defined by
$$
\langle v_d(\lambda x),\varphi(x)\rangle = \frac{1}{|\lambda|^d}\langle v_d(x),\varphi(x/\lambda)\rangle
$$
using Equation (1) with $\varphi(x)$ replaced by $\varphi(x/\lambda)$. Doing the change of variable $x/\lambda \to x$ yields
$$
v_d(\lambda\,\cdot) = \frac{1}{|\lambda|^d}\,v_d + \omega_d\frac{\ln(\lambda)}{|\lambda|^d}\,\delta_0.
$$
Therefore, we can now use the scaling properties of the Fourier transform to get for any $r=1/\lambda>0$
$$
\begin{align*}
(\mathcal{F}(v_d))(r\tilde{x}) &= r^{-d} (\mathcal{F}(v_d(y/r)))(\tilde{x}) 
\newline&= \mathcal{F}(v_d-\omega_d\ln(r)\delta_0)(\tilde{x})
\newline&= \mathcal{F}(v_d)(\tilde{x}) -\omega_d\ln(r).
\end{align*}
$$
Taking $\tilde{x} = \frac{x}{|x|}$ and $r=|x|$ gives
$$
\boxed{\mathcal{F}(v_d)(x) = C_d -\omega_d\ln(|x|)}
$$
where $C_d = \mathcal{F}(v_d)(\tilde{x})$ is a constant since the Fourier transform of a radial function is radial.

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If you want to know the constant $C_d$, the usual trick is to multiply by a Gaussian and use the fact that we know the Fourier transform of a Gaussian. Here remark first that by the Fourier inversion theorem we have
$$
\mathcal{F}(\ln(|x|)) = C_d\, \delta_0 - \frac{v_d}{\omega_d}
$$
Therefore
$$
\begin{align*}
	C_d - \langle\mathcal{F}(\ln(|x|)), e^{-\pi|x|^2}\rangle &= \frac{1}{\omega_d}\langle v_d, e^{-\pi|x|^2}\rangle
	\newline
	&= \int_0^1 \frac{e^{-\pi r^2}-1}{r}\,\mathrm{d} r + \int_1^\infty \frac{e^{-\pi r^2}}{r}\,\mathrm{d} r
	\newline
	&= \int_0^\pi \frac{e^{-t}-1}{2\,t}\,\mathrm{d} t + \int_\pi^\infty \frac{e^{-t}}{2\,t}\,\mathrm{d} t
	\newline
	&= \frac{-\ln(\pi)}{2} + \int_0^\pi \frac{\ln(t)\,e^{-t}}{2}\,\mathrm{d} t + \int_\pi^\infty \frac{\ln(t)\,e^{-t}}{2}\,\mathrm{d} t
	\newline
	&= \frac{-\gamma - \ln(\pi)}{2}
\end{align*}
$$
where $\gamma = -\Gamma'(1) = -\psi(1)$ and I did a polar change of variable and the change $t = πr^2$ and then integrations by parts. But since $\mathcal{F}(e^{-\pi|x|^2}) = e^{-\pi|x|^2}$, we can also compute
$$
\begin{align*}
	\langle\mathcal{F}(\ln(|x|)), e^{-\pi|x|^2}\rangle &= \int_{\mathbb{R}^d} \ln(|x|)\, e^{-\pi|x|^2}\,\mathrm{d} x
	\newline
	&= \omega_d \int_0^\infty \ln(r)\,e^{-\pi r^2} r^{d-1}\,\mathrm{d} r
	\newline
	&= \frac{1}{2\,\Gamma(d/2)} \int_0^\infty (\ln(t)-\ln(\pi))\,e^{-t}\, t^{d/2-1}\,\mathrm{d} t
	\newline
	&= \frac{1}{2} \,\big(\psi(d/2) - \ln(\pi)\big).
\end{align*}
$$
with the same changes of variable. We deduce that $C_d = \frac{\psi(d/2)-\gamma}{2} - \ln(\pi)$. Tell me if you spot any errors!