$R \cong R[[x]]$ 
Is it possible for a commutative ring with 1 to be isomorphic to the
  power series ring over itself? $$R \cong R[[x]].$$ Bonus marks if you
  can answer for $n$ variables more generally, $R \cong R[[x_1, \dots, x_n]]$.

Obviously, $R$ would have to be huge (infinite Krull dimension). It's not clear to me that taking a colimit of iterates of the functor $S \mapsto S[[x_1, \dots, x_n]]$ produces a ring of the desired kind, essentially because 
$$R[[x]][[y]] \not \cong R[[x,y]].$$
So you can't make a construction analogous to what's done for $R \cong R[X]$.
EDIT: Looks like I was under a misapprehension about the above-stated non-isomorphism. Thank you to several people for correcting this important mistake!
 A: As I said in the comments, the zero ring is a first example. Another example is $A[[X_1,X_2,...]]$ where $A$ is a commutative ring with one and the $X_i$ are indeterminates.
Edit. Here is a brief reminder.
If $A$ is a commutative ring with one, if $(X_i)_{i\in I}$ is a family of indeterminates, and if $M(I)\subset A[(X_i)_{i\in I}]$ is the set of monomials, then $A[[(X_i)_{i\in I}]]$ is the set of formal expressions of the form 
$$
\sum_{m\in M(I)}a_mm,
$$ 
with $a_m$ in $A$, equipped with its natural ring structure.
If $I=J\sqcup K$ is a partition of $I$, then the natural morphism 
$$
\left(A[[(X_j)_{j\in J}]]\right)[[(X_k)_{k\in K}]]\to A[[(X_i)_{i\in I}]]
$$ 
is easily seen to be an isomorphism.
A: Let $M$ be a monoid which is "locally finite" by which we mean, for all $m\in M$, $\{(a,b)\in M\times M : ab =m \}$ is finite.
For such a monoid, for a ring $R$, we can make $R^M$ into a ring with the convolution product instead of $R^{\oplus M}$. Lets denote $R^M$ with the convolution product by $R[[M]]$. Now $R[[x]]=R[[\Bbb{N}]]$. Now observe that as sets, by the usual product-hom adjunction, $(R^M)^N = R^{M\times N}$. Moreover, if $M$ and $N$ are locally finite, then $M\times N$ is locally finite. Indeed, one can check that the preserves sums and products, so $R[[M]][[N]]\cong R[[M\times N]]$. Thus we just need to find a locally finite monoid $M$ such that $M\times \Bbb{N}\cong M$. An easy choice is 
$$M=\{u\in \Bbb{N}^\Bbb{N}:u_n\ne 0\text{ for only finitely many $n$}\}.$$
This is intuitively probably the monoid being referred to when people talk about $R[[x_1,x_2,\ldots]]$.
Thus this gives a proof outline that when $A=R[[x_1,x_2,\ldots]]$, then $A[[y]]\cong A$.
