Since
$
\partial_x^2 G
= \nabla\cdot(\partial_xG,0,0)
= \nabla\cdot(\partial_xG \,\hat{x})
$
we get
$$
\iiint_{\Omega} \partial_x^2 G \, dV
= \iiint_{\Omega} \nabla\cdot(\partial_xG \,\hat{x}) \, dV
= \iint_{\partial\Omega} (\partial_xG \,\hat{x}) \cdot n\,dS
= \hat{x}\cdot\iint_{\partial\Omega} \partial_xG \, n\,dS
.
$$
Taking $\Omega=B_r(\mathbf{0})$ makes
$$
\iint_{\partial\Omega} \partial_xG \, n\,dS
= - \iint_{S_r(\mathbf{0})} \frac{x}{|\mathbf{r}|^3} \frac{\mathbf{r}}{|\mathbf{r}|} \, dS
= - \iint_{S_r(\mathbf{0})} \frac{(x^2,xy,xz)}{|\mathbf{r}|^4} \, dS
.
$$
Here, $\iint_{S_r(\mathbf{0})} \frac{xy}{|\mathbf{r}|^4} \, dS$ and $\iint_{S_r(\mathbf{0})} \frac{xz}{|\mathbf{r}|^4} \, dS$ will vanish because of the integrands are odd and the domain of integration is symmetric. So only $\iint_{S_r(\mathbf{0})} \frac{x^2}{|\mathbf{r}|^4} \, dS$ will contribute. Changing to spherical coordinates we get
$$
\iint_{S_r(\mathbf{0})} \frac{x^2}{|\mathbf{r}|^4} \, dS
= \int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \, r^2 \sin\theta \frac{r^2 \sin^2\theta \cos^2\phi}{r^4}
= \int_{0}^{2\pi} d\phi \cos^2\phi \int_{0}^{\pi} d\theta \, \sin^3\theta
= \pi \cdot \frac{4}{3} = \frac{4\pi}{3}
.
$$
Thus we have
$
\iiint_{B_r(\mathbf{0})} \partial_x^2 G \, dV = -\frac{4\pi}{3}
$
independent of $r$ so we have a term $-\frac{4\pi}{3}\delta(\mathbf{r})$ in $\partial_x^2 G.$
But we should also remember that outside of origin, we have
$$
\partial_x^2 G
= \partial_x \left(\partial_x \frac{1}{|\mathbf{r}|} \right)
= \partial_x \left(- \frac{x}{|\mathbf{r}|^3} \right)
= -\frac{1\cdot|\mathbf{r}|^3 - x\cdot 3|\mathbf{r}|^2\cdot x/|\mathbf{r}|}{|\mathbf{r}|^6} \\
= -\frac{|\mathbf{r}|^4 - 3|\mathbf{r}|^2x^2}{|\mathbf{r}|^7}
= -\frac{|\mathbf{r}|^2 - 3x^2}{|\mathbf{r}|^5}
= -\frac{-2x^2+y^2+z^2}{|\mathbf{r}|^5}
.
$$
Without having made certain that nothing has been missed, I therefore claim that
$$
\partial_x^2 G(\mathbf{r})
= \frac{2x^2-y^2-z^2}{|\mathbf{r}|^5} - \frac{4\pi}{3}\delta(\mathbf{r})
.
$$
