This is to give an alternative solution, which works for three general points of the cube and does not involve lumping.

Let $F_{xy}(z)$ be the probability that the random walk started form $z$ visits $x$ before $y$. Let $P$ be the transition matrix of the walk. Then, by conditioning on the first step of the walk, we see that $F_{xy}$ is a solution to the Dirichlet problem $F_{xy}(x)=1,$ $F_{xy}(y)=0$ and 
$$
\sum_w P(z,w)F_{xy}(w)=F_{xy}(z),\quad z\neq x,y.
$$
In fact, it is determined uniquely by these conditions, as the difference of two solutions cannot have a maximum or minimum at $z\neq x,y$. In fact, we can instead solve the following problem: 
\begin{equation}
\Delta G=\delta_x-\delta_y, 
\tag{1}
\end{equation}
where $\Delta=I-P$. Indeed, $\sum_w(\Delta H)(w)=0$ for any $H$, thus, $(\Delta F_{xy})(x)=-(\Delta F_{xy})(y)$, i. e., $\Delta F_{xy}=\alpha(\delta_x-\delta_y)$ for some $\alpha\neq 0$. The solution to (1) is uniquely determined up to an additive constant, so, if $G$ is any such solution, then 
$$
F_{xy}(z)=\frac{G(z)-G(0)}{G(1)-G(0)}.
$$
So far, this is all true for any reversible Markov chain. To solve (1) for the hypercube, we use that the hypercube is an Abelian group and that $\Delta$ commutes with the group action. Concretely, we use the Fourier-Walsh transform. For the hypercube $Q=\{a=(a_1,\dots,a_n):a_i\in\{0,1\}\}$, we have the orthonormal basis of $L^2(Q)$ indexed by $S\subset\{a_1,\dots,a_n\}$ given by 
$$
\xi_S(a)=2^{-\frac{n}{2}}\prod_{j\in S} (-1)^{a_j}.
$$
So, we can decompose $G$ in this basis, $G=\sum_S \hat{G}_S \xi_S$. We compute 
$$
\Delta \xi_S=\left(\frac{1}{n}\sum_{i=1}^n(1-(-1)^{\mathbf{1}_{i\in S}})\right)\xi_S=\frac{2|S|}{n}\xi_S,
$$
while
$$
\hat{(\delta_x)}_S=\sum_{a\in Q}\xi_S(a)\delta_x(a)=\xi_S(x).
$$
Therefore, the equation (1) after the Fourier transform becomes 
$$
\Delta G=\sum_S\frac{2|S|}{n} \hat{G}_S\xi_S=\sum_S(\xi_S(x)-\xi_S(y))\xi_S.
$$
So the solution to (1), up to an additive constant, is given by 
$$
G(z)=\sum_{S\neq \emptyset}\frac{n}{2|S|}(\xi_S(x)-\xi_S(y))\xi_S(z).
$$