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The number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\ -i \ \ i \ -i \ \cdots\ i \ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\biggl(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\biggr)^{n} = \left(\frac{e^x + e^{-x}}{2}\right)^n. $$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula.