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 $b_t\supset[t-a,t+a]\times B^{d-1}_a$, 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} dx_1\,e^{-x_1^2/2} \;C_{d,a}
	\ge c_d \frac{e^{-(t-a)^2/2}}{t}
\end{equation}
eventually (that is, for all large enough $t>0$), where $c_d:=\frac{C_{d,a}}{2\sqrt{2\pi}}>0$. 

On the other hand, 
\begin{equation}
	\nu(b_t+x_0)\le\frac1{\sqrt{2\pi}}\int_{t+2a-2a}^\infty dx_1\,e^{-x_1^2/2}\le e^{-t^2/2}.
\end{equation}

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