= Solution
A complete excursion from $B$ along any one arm and back to $B$ has mean length $L=2n$. This follows from the same path recurrence, or from <Kac's lemma>: the graph has $3n$ edges, the <stationary distribution> of simple random walk assigns mass $\deg(B)/(2|E|)=3/(6n)=1/(2n)$ to $B$, so its mean return time is $L=2n$.
Let $x$ be the mean time to hit $E$ starting at $B$. On each visit to $B$, the first step reaches $E$ with probability $1/3$; with probability $2/3$ it begins a wrong-arm excursion of mean duration $L$, after which the problem restarts. Thus
$$
x=\frac13+\frac23(L+x),
\qquad
x=1+2L=4n+1.
$$
The walk first takes mean time $n^2$ to reach $B$, so the requested time from $A$ is
$$
\boxed{\mathbb E_A[T_E]=n^2+4n+1=(n+2)^2-3}.
$$
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