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I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.

The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).

As you mentioned an other option is to compute these derivatives:
$$partial_i u =frac{x_i}{|x|^3}$$
so:
$$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
-3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
but none of these functions are in $L^1$. For example using spherical coordinates you obtain
$$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.

The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.