Let us consider the case $n=2$. Assume $p_k$ is at the origin and $p_0$ is at the point $(1,0)$ on the $x$-axis, and that $r<1$.

To find the distribution of $h_1$, by subtracting a constant it suffices to find the distribution of $d:=d(p_1,0)$.

The density for $d$, $f_d(R)$, is proportional to $R\Theta$ where $\Theta$ is length of the interval of angles $\theta$ for which the point $(R\cos\theta,R\sin\theta)$ is within $r$ of $(1,0)$. Now, calculation shows that
$$
|(R\cos\theta,R\sin\theta)-(1,0)| < r
$$
is equivalent to
$$
|\theta| < \arccos\left(\frac{R^2+1-r^2}{2R}\right)
$$
so $f_d(R)$ is proportional to
$$
R\arccos\left(\frac{R^2+1-r^2}{2R}\right),\quad R\in [1-r,1+r].
$$
and $f_d(R)=0$ for $R\not\in [1-r,1+r]$.