For $n=3$ this is easy.  A triple is in $A$ iff its maximum and minimum are within $d/2$.  So  $A \cap B_p$ is non-empty iff the maximum and minimum of $p$ are within $c=\min(1,d/2+2r)$.  This has probability
$$6\left(\int_{x=0}^{1-c}
\int_{y=x}^{x+c}
\int_{z=y}^{x+c}
dz\, dy\, dx
+
\int_{x=1-c}^1
\int_{y=x}^1
\int_{z=y}^1
dz\, dy\, ds
\right)
$$
$$=6((1-c)c^2+c^3/6)=6c^2-5c^3.$$

We'd need other descriptions of $A$ in higher dimensions to make this work for larger $n$.