Of course, the charge distribution is not ##0## everywhere. It rather has a singularity at the location of the dipole.
It's well worth to consider this in some detail. So let's start from the somewhat simpler problem of point charge, which we put in the origin of our coordinate system. Also there ##\rho(\vec{r})=0## everywhere except at ##\vec{r}=0##, where the charge density must be singular. Formally that's written in terms of the Dirac distribution (where "distribution" here means "generalized function"), i.e.,
$$\rho(\vec{r})=q \delta^{(3)}(\vec{r}).$$
Now consider a dipole. Intuitively you can think about it as having two charges ##q## and ##-q##. Let's put the negative one at ##\vec{r}=0## and the positive one at some place ##\vec{d}##. Then the charge distribution is
$$\rho(\vec{r}) = q [\delta(\vec{r}-\vec{d})-\delta(\vec{r})]=-q \vec{d} \cdot \vec{\nabla} \delta(\vec{r}) + \mathcal{O}(q |\vec{d}|^2).$$
Now we consider the limit ##|\vec{d}| \rightarrow 0## such that ##\vec{p} =q \vec{d}=\text{const}##. Then the contributions to higher order vanish, and you get
$$\rho(\vec{r})=-\vec{p} \cdot \vec{\nabla} \delta^{(3)}(\vec{r}).$$
The electrostatic potential becomes
$$\Phi(\vec{r})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|} =- \int_{\mathbb{R}^3} \frac{1}{4 \pi |\vec{r}-\vec{r}'|} \vec{p} \cdot \vec{\nabla}' \delta^{(3)}(\vec{r}').$$
Integration by parts gives
$$\Phi(\vec{r})=+ \frac{1}{4 \pi \epsilon_0} \int_{\mathbb{R}^3} \mathrm{d}^3 r' \vec{p} \cdot \vec{\nabla}' \frac{1}{|\vec{r}-\vec{r}'|} =\frac{\vec{p} \cdot \vec{r}}{|4 \pi \epsilon_0 r^3},$$
i.e., the dipole field.
So you get
$$\Delta \frac{\vec{p} \cdot \vec{r}}{4 \pi \epsilon_0 r^3}=-\frac{1}{\epsilon}_0 \rho(\vec{r})=\frac{1}{\epsilon_0} \vec{p} \cdot \vec{\nabla} \delta^{(3)}(\vec{r}).$$