The first examples of closed 3-manifolds not admitting a conformally flat (Mobius) structure [were given by Goldma][1]n: 3-manifolds modeled on the Nil geometry (this link was given by Macbeth in the comments above). Sol manifolds also don't have a Mobius structure, whereas 3-manifolds modeled on the other six geometries do. [Misha Kapovich][2] has many results on conformally flat structures. In his thesis, he shows that certain classes of Haken 3-manifolds have finite-sheeted covers which are (uniformizable) conformally flat, and he shows there is a graph manifold which has no conformally flat structure, but which has a finite-sheeted cover which does. Kulkarni showed that connect sums of conformally flat manifolds are conformally flat. On the other hand, Kapovich's student Hwang showed that for any 3-manifold $M$, there is a 3-manifold $N$ such that $M {\\#} N$ is conformally flat. I don't know of any recent activity on the topic.

As for your question on maximal Lie group actions on 3-manifolds, I've heard this before too, but I don't know a reference. I think you can prove it by analyzing the action of the isotropy group of a point on the jet space at that point.

Addendum: I had a discussion with Cooper about the last question (maximal Lie groups acting faithfully on an $n$-manifold), and we have an idea how to approach it (at least for smooth actions). If a Lie group $G$ acts smoothly and faithfully on a manifold $M$, then one obtains a homomorphism 
$\Phi: G \to Diff(M)$. Then we get a map $\phi:g \to Vect(M)$, where $g$ is the Lie algebra of $G$, and $Vect(M)$ is the Lie algebra  of $Diff(M)$. So one wants to classify maximal finite-dimensional Lie algebras of $Vect(M)$. First, if $g$ is not semisimple, then it has a non-trivial center $c$, which is generated by a non-zero smooth vector field $V$. One should then be able to take a quotient $M/V$ of $M$ with action on $M/V$ by the Lie algebra $g/c$ and apply an inductive argument. Actually, one should try to do this only locally, since the quotient might not be nice. Then assume $g$ is semisimple. Its Cartan subalgebra gives $R^m$ acting on $R^n$. This gives $m$ commuting vector fields, and in particular gives $m$-dimensional coordinates 
at a point, so $m\leq n$. Now, one needs to appeal to the classification of semisimple Lie algebras to finish off the proof (the Cartan subalgebra together with Weyl group determines the Lie algebra), and then apply the inductive argument to deal with the radical of the Lie algebra. I haven't worked out how to do this, but it seems like a plausible approach. I suspect an argument like this may be well-known in the right circles. 


  [1]: http://www.ams.org/mathscinet-getitem?mr=701512
  [2]: http://www.math.ucdavis.edu/~kapovich/