Let us first consider the function $g_n(k)=\left|\frac{\hat{f}(k)}{||k||^2} \right| \cdot \mathbb 1_{A_n}(k)$ where $A_n= B(0,n) \cap B(0,1/n)^c$.
Note that $g_n$ is non-negative, integrable (why?), increasing with respect to $n \in \mathbb N$, and that, almost surely, $g_n \to_n \left|\frac{\hat{f}(.)}{||.||^2} \right|$. Hence $$\int g_n(k)dk \to \int{ \left|\frac{\hat{f}(k)}{||k||^2}\right|dk}$$ by the monotone convergence theorem.
Now note that $g_n$ is rotation invariant (why?), hence for any $t>0$ it satisfies $\forall x_0 \in \partial B(0,t)$ we have $\frac{1}{|\partial B(0,t)|}\int_{\partial B(0,t)}g_n(x)dx = g(x_0)$. (Reminder: $\partial B(0,t)$ is the sphere of radius $t$ and $S^2= \partial B(0,1)$ ).
In what will follow, I will choose the point $(1,0,0)=x_0$, but any other point on $S^2$ can do just as well.
Let us now integrate on the spherical level sets: $$\int g_n(k)dk=\int_{0}^{\infty}\int_{S^2}r^2g_n(rw)d\sigma_{S^2}(w)dr= \frac{1}{|S^2|}\cdot\int_{0}^{\infty} r^2 g_n(r \cdot (1,0,0))dr= \\ \frac{1}{|S^2|}\cdot\int_{0}^{\infty} r^2 g_n(r \cdot (1,0,0))dr.$$
We will now split the integral: $$\int_0^1 + \int_1^\infty .$$ For the first integral, we use the fact that $r^2 g_n(r \cdot (1,0,0))= r^2 \frac{|\hat{f}(r,0,0)|}{r^2}\mathbb1_{[1/n,n]}=|\hat{f}(r,0,0)| \mathbb1_{[1/n,n]}$ which is bounded by some constant $C$ independent of $n$.
For the second integral, we apply the formula that you provided, namely $r^2g_n(r,0,0)=r^2\mathbb1_{[1/n,n]} \cdot \left| \frac{\sin(2\pi r)-2\pi r\cos(2\pi r)}{2\pi^2r^{2+3}}\right| \le r^2 \mathbb1_{[1/n,n]} \cdot \left( \frac{|\sin(2\pi r)|}{2\pi^2r^{5}}+\frac{|2\pi r\cos(2\pi r)|}{2\pi^2r^{5}}\right) \\ \le \mathbb1_{[1/n,n]} \cdot \left(\frac{c_1}{r^3} + \frac{c_2}{r^2}\right)$.
Can you take it from here and show that $\int g_n(k)dk$ is a bounded increasing sequence, hence that it converges?
Here is an alternative proof using the fact that in $\mathbb R^d$,
$||x||^{-r}$ is integrable on $B(0,1)$ iff $r<d$
and
$||x||^{-r}$ is integrable on $B(0,1)^c$ iff $r>d$
In your case, $d=3$. Note that $ \mathbb 1_{B(0,1)}|\frac{\hat{f}(k)}{||k||^2}| \le \mathbb 1_{B(0,1)}|\frac{\hat{f}(0)}{||k||^2}|$ which is integrable.
Also note that $\mathbb 1_{B(0,1)^c}|\frac{\hat{f}(k)}{||k||^2}|= \mathbb 1_{B(0,1)^c} \left| \frac{\sin(2\pi ||k||)-2\pi r\cos(2\pi ||k||)}{2\pi^2||k||^{2+3}}\right| \le \mathbb 1_{B(0,1)^c} \left(\frac{c_1}{||k||^5} + \frac{c_2}{||k||^4}\right)$ which is also integrable.
Now you can conclude using the inequalities.