Which lattices are quotients of finite powerset lattices? Let $S$ be a finite set, and let $2^S$ be its powerset, regarded as a lattice. Let $L$ be a quotient (in the category of lattices and maps which preserve $\top,\bot,\wedge,\vee$) of $S$. What can we say about $L$?
 In fact, what I'd really like to know is: which finite semilattices are retracts (via $\bot,\vee$-preserving maps) of finite powerset lattices. I think the two questions are equivalent via a short-but-not-immediate argument.
Note that if $P$ is an arbitrary finite poset, then the lattice $2^P$ of poset maps $P \to 2$ is an example of such a lattice.
 A: Tim-
Emil answered the question you asked, but since you wrote ``what I'd really like to know is: which finite semilattices are retracts (via $\bot$, $\vee$-preserving maps) of finite powerset lattices'' let me add to his answer.
The $2$-element semilattice is injective in the class of semilattices. 
The class of injectives is closed under products and retracts.
Up to isomorphism, the powers of the $2$-element semilattice are the power-set semilattices. Hence retracts of power-set semilattices must be injective.
Conversely, since the $2$-element semilattice is the only subdirectly irreducible semilattice, every semilattice is embeddable in some power $2^S$. And since an injective is a retract of any extension, it follows that every injective arises as a retract of some $2^S$,
Thus, the retracts of the power-set semilattices, $2^S$, are exactly the injective semilattices.
Theorem 2.8 of 
The Category of Semilattices 
ALFRED HORN and NAOKI KIMURA
Algebra universalis 1 (1971), 26-38.
proves that a (meet-)semilattice is injective iff it is complete and satisfies an infinite distributive law, namely that the meet distributes over infinite joins.
It follows that a finite semilattice is a retract of a power-set semilattice iff it is the semilattice reduct of a finite distributive lattice. (This shows that the examples in your last struckout paragraph exhaust all examples.)
A: The class of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are exactly the lattice reducts of finite Boolean algebras.
In particular, any quotient $L$ of $2^S$ has to be a bounded distributive lattice, as the class of distributive lattices is a variety. Moreover, if $x\in L$ is an image of a set $A\subseteq S$, then the image of $S\smallsetminus A$ is an element $y$ such that $x\lor y=\top$ and $x\land y=\bot$. Thus, $L$ is complemented, i.e., it is a Boolean algebra, and as such it is isomorphic to $2^{S'}$ for some set $S'$.
