I know that these things are better calculated using spherical coordinates and/or Fourier transforms. I tried to calculate this in cartesian coordinates anyway:
Let $G(\mathbf{x},\mathbf{x_0}) = \frac{1}{|\mathbf{x}-\mathbf{x_0}|}$ be the fundamental solution for the Laplace operator $\Delta$ in $\mathbb{R}^n$ for $n= 3$.
This means that $\Delta_{\mathbf{x}} G(\mathbf{x},\mathbf{x_0}) = \delta(\mathbf{x}-\mathbf{x_0})$, which should be $0$ everywhere except at $\mathbf{x}=\mathbf{x_0}$ (where it must be treated as a distribution instead of a normal function).
Let $\mathbf{x_0} = 0$ for simplicity, then:
$$ \begin{align} \Delta G(\mathbf{x}) & = \nabla^2 \frac{1}{|\mathbf{x}|} \\ & = \nabla \left(-\frac{1}{\mathbf{|x|}^3}\mathbf{x}\right) \\ & = \frac{2}{|\mathbf{x}|^3} \end{align} $$
This is a strictly positive function everywhere except at $\mathbf{x}=0$, so it clearly is not equal to $0$ (even in the sense of distributions).
What am I doing wrong?
Your last line is wrong. $$\Delta G(x)=\nabla\cdot \nabla G(x)=\nabla\cdot \nabla \frac{1}{|x|}$$ $$=\nabla\cdot \left(-\frac{\vec{x}}{|x|^3}\right)$$ And the divergence of $-\frac{\vec{x}}{|x|^3}$ is equal to $0$. You just have to compute it carefully:
$$\frac{\partial}{\partial x_i}\left(-\frac{x_i}{|x|^3}\right)=-\frac{(x_i)'\cdot |x|^3-x_i\cdot (|x|^3)'}{|x|^6}=-\frac{|x|^3-3x_i|x|^2\left(\frac{1}{2|x|}\right)2x_i}{|x|^6}$$ $$=-\frac{|x|^3-3x_i^2|x|}{|x|^6}$$ Then the divergence is $$\nabla\cdot\left(-\frac{\vec{x}}{|x|^3}\right)=\sum\limits_{i=1}^{3}{\frac{\partial}{\partial x_i}\left(-\frac{x_i}{|x|^3}\right)}=\sum\limits_{i=1}^{3}{-\frac{|x|^3-3x_i^2|x|}{|x|^6}}=0$$
Note that this is true only for $n=3$.