Why wasn't Bertrand Russell surprised by the set of all sets that contain themselves? Russell's paradox deals with the question: "Does the set of all sets that do not contain themselves, contain itself?"
What about the question: "Does the set of all sets that contain themselves, contain itself?"
I find the latter an equally interesting and contradictory question/statement. 
Doesn't the set of all sets that contain themselves imply that for every such set $P$  you have to create an additional set $Q$ containing $P$ and all the sets in $P$, this process going  on infinitely?

There's no paradox here.
If there are no sets that contain themselves (such as in modern ZFC set theory), then $\{x\mid x\in x\}$ is simply the empty set -- and we know the empty set exists!
If there are sets that contain themselves (but not too many of them), then we can still assert that there is a set that consists of all of them. That set may or may not contain itself; neither choice leads to a contradiction.
On the other hand if you have too many self-containing sets -- such that if we're working with an anti-foundation axiom that guarantees that for every $x$ there is an $y$ such that $y=\{x,y\}$ -- then we can't have a set of all of them, at least not in a ZF background, because its union would be a set of all sets, which leads to the ordinary Russell and Cantor paradoxes.

Doesn't the set of all sets that contain themselves imply that for every such set $P$ you have to create an additional set $Q$ containing $P$ and all the sets in $P$, this process going on infinitely?

Well, suppose $P=\{x\mid x\in x\}$ exists. Either $P\in P$ or $P\notin P$.
If $P\in P$, then $Q=P\cup\{P\}$ is just $P$ itself, so nothing new has been produced.
If $P\notin P$, then $Q=P\cup\{P\}$ certainly ought to exist, but it is not a set of all sets that contain themselves, because one of its elements is $P$ and $P$ doesn't contain itself. ($Q$ may or may not contain itself, but that's still not a problem -- it will if and only if it does).
