We have 
\begin{equation*}
	F(x_0)=\infty \tag{1}\label{1}
\end{equation*}
for any nonzero $x_0$. 

Indeed, by spherical symmetry, without loss of generality 
\begin{equation*}
	x_0=(2a,0,\dots,0)
\end{equation*}
for some real $a>0$. 

Let $b_t:=B_{2a}((t,0,\dots,0))$ for real $t>0$. Let 
\begin{equation*}
	C_{d,a}:=\nu_{d-1}\big(B^{d-1}_a\big),
\end{equation*}
where $\nu_k$ is the standard Gaussian measure over $\Bbb R^k$ and $B^k_r$ is the ball in $\Bbb R^k$ of radius $r$ centered at the origin. 
Note that $b_t\supset[t-a,t-a/2]\times B^{d-1}_a$, if $\Bbb R^d$ is identified with $\Bbb R\times\Bbb R^{d-1}$. So, 
\begin{equation*}
	\nu(b_t)\ge\frac1{\sqrt{2\pi}}\int_{t-a}^{t-a/2} dx_1\,e^{-x_1^2/2} \;C_{d,a}
	\ge c_d\, e^{-(t-a/2)^2/2},
\end{equation*}
where $c_d:=\frac{C_{d,a}}{\sqrt{2\pi}}\frac a2>0$. 

On the other hand, 
\begin{equation*}
	\nu(b_t+x_0)\le\nu([t+2a-2a,\infty)\times\Bbb R^{d-1})=
	\frac1{\sqrt{2\pi}}\int_t^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}.
\end{equation*}

So, 
\begin{equation*}
	\frac{\nu(b_t)}{\nu(b_t+x_0)}\ge c_d\,\frac{e^{-(t-a)^2/2}}{e^{-t^2/2}}
	\to\infty
\end{equation*}
as $t\to\infty$. So, \eqref{1} is proved. $\quad\Box$