Proving random walk on $Z^3$ is transient using network theory

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I am looking at Example 21.9 in Page 295 here which uses flows of electrical network theory to show that $Z^3$ is transient and I have a couple of questions about the proof.

Proof: To each directed edge $\vec{e}$ in the lattice $Z^3$, attach an orthogonal unit square $\square_e$ intersecting $\vec{e}$ at its midpoint $m_e$. Let $\sigma_e$ be the sign of the scalar product between $\vec{e}$ and the vector from $0$ to $m_e$. Define the flow along $\vec{e}$, namely $\theta(\vec{e})$, to be the area of the radial projection of $\square_e$ onto the sphere of radius $1/4$ centered at the origin, multiplied by $\sigma_e$. By considering the projections of all faces of the unit cube centered at a lattice point $x\neq 0$, we can verify that $\theta$ satisfies the node law at $x$. (Almost every ray from the origin that intersects the cube enters it through a face $\square_{xy}$ with $\sigma_{xy}=-1$ and exits through a face $\square_{xz}$ with $\sigma_{xz} = 1$. Note that $\theta(\vec{0y})>0$ for all neighbors $y$ of $0$. Hence $\theta$ is a nonzero flow from $0$ to $\infty$ in $Z^3$. For a square of distance $r$ from the origin, projecting onto the sphere of radius $1/4$ reduces area by order $r^2$. Thus:

$$\mathcal{E}(\theta) \leq \sum_{n}C_1 n^2\left(\frac{C_2}{n^2}\right)^2 < \infty$$

and $Z^3$ is transient. ($\mathcal{E}(\theta)$ is the energy of a flow, $\mathcal{E}(\theta) := \sum_e \theta(e)^2 r(e)$, $r(e)$ is the resistance on an edge).

Questions: 1) Let $x$ be on $X-$axis, then I understand why $\sigma_{xy}=-1$ and $\sigma_{xz}=1$ but the area of projections of squares on edges $xy$ and $xz$ would be different because the square on $xz$ is farther from the origin. So I am not sure how node law is satisfied?

2) Not entirely sure how we got $$\mathcal{E}(\theta) \leq \sum_{n}C_1 n^2\left(\frac{C_2}{n^2}\right)^2 $$ I believe we are looking at a cube of $n$ edges in one dimension. Then the distance from origin to any square would be order $n$ and hence I understand why we have $\frac{C_2}{n^2}$ term but not sure why we have $C_1 n^2$ term. If each edge has resistance $=1$, the expression means there are order $n^2$ edges which is not true as there are more edges than that?