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 $$6\left(\int_{x=0}^{1-c} \int_{y=x}^{x+c} \int_{z=y}^{x+c} dz\, dy\, dx + \int_{x=1-c}^1 \int_{y=x}^1 \int_{z=y}^1 dz\, dy\, dx \right) $$ $$=6\left(\frac{c^2-c^3}{2}+\frac{c^3}{6}\right)=3c^2-2c^3.$$

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

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 coordinates of $p$ are within $c=\min(1,d/2+2r)$. This has probability $$6\left(\int_{x=0}^{1-c} \int_{y=x}^{x+c} \int_{z=y}^{x+c} dz\, dy\, dx + \int_{x=1-c}^1 \int_{y=x}^1 \int_{z=y}^1 dz\, dy\, dx \right) $$ $$=6\left(\frac{c^2-c^3}{2}+\frac{c^3}{6}\right)=3c^2-2c^3.$$

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

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 $n$.

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 $$6\left(\int_{x=0}^{1-c} \int_{y=x}^{x+c} \int_{z=y}^{x+c} dz\, dy\, dx + \int_{x=1-c}^1 \int_{y=x}^1 \int_{z=y}^1 dz\, dy\, dx \right) $$ $$=6\left(\frac{c^2-c^3}{2}+\frac{c^3}{6}\right)=3c^2-2c^3.$$

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