In fact $S_d$ does not converge to zero when $d\to\infty$, at least if $a_x=1$ for every $x$. Here is a proof. 

For every $x$, $G_d(x)=P_0^{(d)}(\mathtt{hit}\ x)G_d(0)\le G_d(0)$ hence $G_d(x)^2/G_d(0)\le G_d(0)$. For every $z$ in $[0,G_d(0)]$, $\sqrt{1+z}-1\ge c_dz$ with 
$$
c_d=\frac1{G_d(0)}[\sqrt{1+G_d(0)}-1]=\frac1{\sqrt{1+G_d(0)}+1}.
$$
Hence, for every $(a_x)$,
$$
S_d\ge c_dT_d,\quad\mbox{with}\
T_d=\sum_xa_xG_d(x)^2/G_d(0).
$$
In the special case where $a_x=1$ for every $x$, by reversibility,
$$
\sum_xG_d(x)^2=\sum_{i,j}\sum_xp_i^{(d)}(0,x)p_j^{(d)}(x,0)=\sum_{i,j}p_{i+j}^{(d)}(0,0)=\sum_{i}(i+1)p_i^{(d)}(0,0),
$$
hence
$$
\sum_xG_d(x)^2>\sum_{i}p_i^{(d)}(0,0)=G_d(0).
$$
In particular, $T_d>1$. Now $G_d(0)$ is a nonincreasing function of $d$ hence $c_d\ge c_5$ for every $d\ge5$ and $S_d>c_5$ for every $d\ge5$.