Is $PH \subseteq PP$?
Huck, as Lance and Robin pointed out, we do have oracles relative to which PH is not in PP.  But that doesn't answer your question, which was what the situation is in the "real" (unrelativized) world!

The short answer is that (as with so much else in complexity theory) we don't know.

But the longer answer is that there are very good reasons to conjecture that indeed PH &sube; PP.

First, Toda's Theorem implies PH &sube; BP.PP, where BP.PP is the complexity class that "is to PP as BPP is to P" (in other words, PP where you can use a randomization to decide which MAJORITY computation you want to perform).  Second, under plausible derandomization hypotheses (similar to the ones that are known to imply P=BPP, by Nisan-Wigderson, Impagliazzo-Wigderson, etc.), we would have PP = BP.PP.

Addendum, to address your other questions:

(1) I'd say that we don't have a compelling intuition either way on the question of whether PP = PPP.  We know, from the results of Beigel-Reingold-Spielman and Fortnow-Reingold, that PP is closed under nonadaptive (truth-table) reductions.  In other words, a P machine that can make parallel queries to a PP oracle is no more powerful than PP itself.  But the fact that these results completely break down for adaptive (non-parallel) queries to the PP oracle suggests that maybe the latter are really more powerful.

(2) Likewise, NPPP and coNPPP might be still more powerful than PPP.  And PPPP might be more powerful still, and so on.  The sequence P, PP, PPP, PPPP, PPP^PP, etc. is called the counting hierarchy, and just as people conjecture that PH is infinite, so too one can conjecture (though maybe with less confidence!) that CH is infinite.  This is closely related to the belief that, in constant-depth threshold circuits (i.e., neural networks), adding more layers of threshold gates gives you more computational power.