Does anyone know of a global proof (involving no local argument) of Serre Duality at the level of varieties or manifolds (as opposed to schemes).

$\begingroup$ Could you clarify what "no local argument" means? Is the Hodge theorem a "local argument" because being harmonic is a local condition? $\endgroup$ – Ben Webster♦ Nov 19 '09 at 15:47
You might like the proof in section 5.3 of Voisin's book Hodge theory and complex algebraic geometry.
Since a $\bar{\partial}$Operator is an elliptic Operator, you can use elliptic theory in order to prove the Serre duality. In fact the Serre duality is a kind of corollary of the"fundamental theorem" (as I Know it). In fact, you do not need the Hodge theorem, since the Hodge theorem itself is a corollary of the theorem.
For a reference of this "fundamental theorem" (perhaps slightly reformulated) I would refer to one of the following:
 Wells, Differential Analysis on Complex Manifolds;
 Gilkey, Invariance Theory, the Heat Equation and the Atiyah Singer Index Theorem
(a complete proof for pseudodifferential operators)
 Warner, Foundations of Differentiable Manifolds
(the theorem is included as an exercise on the last page
 Kazdan, Lecture Notes on Applications of Partial Differential Equations to Some Problems in Differential Geometry (online available here)
(Corollary 2.5, for a sketch of the proof)
I like the presentation from Analytic methods in algebraic geometry by Demailly. Here is the link: http://wwwfourier.ujfgrenoble.fr/~demailly/manuscripts/eem2007.pdf.
I thought I'd offer a hightech alternative for certain varieties. If $X$ is smooth and projective over a field $k$ then Bondal and van den Bergh give a proof here that $D^b(\mathrm{Coh}X)$ is saturated which is a strong representability condition on cohomological/homological functors to the category of $k$ vector spaces. It follows immediately that $D^b(\mathrm{Coh}X)$ has a Serre functor by using the fact that $Hom(A,)^*$ is representable for every bounded complex of coherent sheaves $A$.
Have you looked at the one in Griffiths and Harris? That's at least rather different from the general nonsense one in Hartshorne.