Let $G_0(x)=G(x,0)$ be the Green's function of the simple symmetric random walk on $\mathbb Z^d$, $d\geq 3$. The question is whether $G_0$ must always vary locally, i.e., whether
$$ \sum_{\substack{y\in\mathbb Z^d:\\ |y-x|=1}} |G_0(x)-G_0(y)| >0 $$
holds for all $x\in\mathbb Z^d$.
The claim seems intuitive in view of the random walk interpretation of $G_0$, i.e., $G_0(x)=\mathbb P(\exists n: S_n=x) G_0(0)$.
For large $x$, the claim can be verified from the asymptotics of $G_0$.
We will use the elementary fact that for $m \ge k \ge m/2$, the binomial coefficients satisfy $${m\choose k+1} <{m \choose k}. \quad (\#)$$
The case $x=0$ is obvious so we may assume $x$ has some nonzero coordinate. By symmetry, we may assume that $x_1>0$. Then it suffices to show that for every $y \in \mathbb Z^d$ that satisfies $y_1 \ge 0$, the point $z=z(y)$ that agrees with $y$ in all coordinates except the first, where $z_1=y_1+2$, we have the strict inequality $$P(S_n=z)<P(S_n=y)$$ for all $n$ such that $P(S_n=z)>0$. Let $A_n$ be the (random) set of steps among the first $n$ when the random walk moved in the first coordinate, and let $w^*$ denote the projection of a node $w \in\mathbb Z^d$ to coordinates $2,3 ,\ldots, d$. Fix $y$ with $y_1 \ge 0$ and let $z=z(y)$ as above, so that $z^*=y^*$. If $A_n$ satisfies $$P(S_n=z \,|A_n)>0 \,,$$ then the cardinality $|A_n|$ and $z_1$ must have the same parity, and by $(\#)$, $$P(S_n=z\, | \, A_n)=P(S_n^*=z^* \,|A_n) \cdot {|A_n| \choose \frac{|A_n|+z_1}{2}}2^{-|A_n|} \; < $$ $$ \: P(S_n^*=y^* \,|A_n) \cdot {|A_n| \choose \frac{|A_n|+y_1}{2}}2^{-|A_n|}= P(S_n=y\, | \, A_n) \,.$$ Taking expectations (i.e., averaging over $A_n$) gives $P(S_n=z)<P(S_n=y)$.
If e.g. the random walk can only move in the positive direction relative to an oriented hyperplane $H$ in $\mathbb R^d$, then the Green function will be locally $0$ and hence constant for all points $x\in\mathbb Z^d$ at distance $>1$ from $H$ in the negative direction relative to $H$.
The local zero-ness and hence the local constancy of the Green function will also occur if the random walk can only move along a sufficiently small sublattice of $\mathbb Z^d$.
In light of your remark, it suffices to show the same about $\mathbb P^0(\exists n: S_n=x)$ . Let $x$ be any point away from the origin, and let If $\tau_x = inf \lbrace n: S_n=x \rbrace $. Let $N_x = \lbrace y : |x-y| = 1 \rbrace $. Then conditioning on the first time the walk hits $N_x$ $$$$ $$\mathbb P^0(\tau_x < \infty ) = \Sigma_{y \in N_x} \mathbb P^0(\tau_y < \infty )\mathbb P^y(\tau_x < \infty ) $$ $$ < \Sigma_{y \in N_x} \mathbb P^0(\tau_y < \infty )$$ where the first line follows because you must hit $N_x$ before x and the strict inequality follows because $ P^y(\tau_x < \infty ) < 1 $ for all y.
