Let's fix a finite field F and consider abelian varieties of dimension g over F. Can we say how many isogeny classes there are? Is it even clear that there's more than one isogeny class? For g=1, and some fixed F with characteristic not 2 or 3, we could probably just write down all the Weierstrass equations and count isogeny classes by brute force, but is there a cleaner way to do it?
Let q be the order of your finite field. Then the category of abelian varieties over $\mathbb{F}_q$ up to isogeny is semisimple  any object is isogenous to a product of simple ones in an essentially unique way, so this reduces your question to one about simple objects.
For simple abelian varieties over $\mathbb{F}_q$, there is the TateHonda classification which states that the isogeny classes are in bijective correspondence with Weil $q$integers (algebraic numbers that have absolute value $q^{1/2}$ in all complex embeddings) up to Galois conjugacy.
I learned this from Milne's "Points on Shimura varieties mod $p$" (one of the articles from the Corvallis proceedings that Kevin mentioned), which has a nice and fairly elementary discussion in section 5.

$\begingroup$ This article is "one of the articles before just Langlands' article in Corvallis" mentioned in my answer. I'm sure that Milne's article gets to the heart of the matter at hand. $\endgroup$ – Kevin Buzzard Nov 7 '09 at 13:34

$\begingroup$ Ah, thank you  I've updated accordingly. $\endgroup$ – Tyler Lawson Nov 7 '09 at 13:53

$\begingroup$ "Weil $q$integers (algebraic numbers ..." you mean "algebraic integers? $\endgroup$ – shenghao May 17 '11 at 13:01
Here's a related question on which there has been much work. If you actually need an answer to your question for some other reason and you follow this up, I'd be interested to see where it goes.
Let $X$ be some moduli space of abelian varieties (so, for $g>1$, there will be extra structure beyond what you have mentioned, for example we'll also be considering a polarization of some sort). In some cases (for example, if the polarization is principal) then X will be a Shimura variety, so will be related to a quotient of a reductive group (a symplectic group, for example). In the $g=1$ case we're talking about modular curves. Eichler computed the HasseWeil zeta function of a modular curve in 1954, and one way of phrasing that statement in a more lowlevel way is that he managed to compute the size of $X(k)$ for any finite field $k$ (at least where $X$ has good reduction). His results were generalised by Shimura. These guys used something called the "congruence relation", which boiled down to computing the reduction of some modular curves mod $p$ where $p$ is a prime dividing the level.
Then Ihara came along and did the calculation a different way. He computed $X(k)$ for any finite field $k$ but this time using the trace formula. OhI should say what the answer was: it turns out that the HasseWeil zeta function of the modular curve $X$ is related to $L$functions attached to modular forms.
The reason I mention all of this is that by the 1970s Langlands argued that this result was the tip of the iceberg, and the HasseWeil zeta function of a large class of Shimura varieties should be related to $L$functions of automorphic forms in some huge generality. Langlands' paper is http://www.ams.org/online_bks/pspum332/pspum332ptIV5.pdf and it's sort of incomprehensible. In it he makes a conjecture which in some sense boils down to a formula for $X(k)$ for $X$ running through a large class of Shimura varieties. But if you look at that link, and then remove the last bit (the name of the pdf file) you'll find the web pages of a book ("Corvallis") and the other articles just before Langlands' offer more downtoearth explanations of special cases of how to compute these numbers: the conjectural formulae of Langlands are to a large extent established for certain Hilbert modular varieties, for example.
So why do I mention all this? Well, the reason is that if you actually look at the proofs of these counting formulas, then, once your head stops hurting, you see that in fact the strategy for counting these points is precisely what you're hinting at above: you split $X(k)$ up into isogeny classes, and then count the size of each isogeny class, and then add them up. And even though the conjecture of Langlands has not been proved in full, the strategy still applies and gives a very powerful handle on the question you're asking.
So, for example, if you were to apply Langlands' ideas (or, more practically, the expositions of them in the papers preceding them) to the case of elliptic curves over $Q$ you would see a concrete machine made for counting isogeny classes of elliptic curves over finite fields. As I said, anyone who did this would learn a lot more about the methods of these papers than I currently understand and they might want to explain what happens.
Jim Milne has been reading this forum recently, and he is one of the experts in this area, so he will probably be able to give you a better answer, in the sense that he'd be able to give you some feeling as to what would happen were you to follow through Langlands' counting programme in the symplectic case.

$\begingroup$ I'll take a look at those articles, thanks. $\endgroup$ – Rebecca Bellovin Nov 8 '09 at 10:13
Let F_{q} be a finite field. Two elliptic curves E_{1} and E_{2} defined over F_{q} are F_{q}rationally isogenous iff # E_{1}(F_{q}) = # E_{2}(F_{q}).
(==>) Let \varphi be an isogeny, take l a prime which is prime to q*(degree of \varphi). Then \varphi induces a Frobeniusequivariant isomorphism on the ladic Tate modules, so the characteristic polynomials of Frobenius agree. Since
E(F_{q}) = q+1  trace(Frob),
this implies that the two elliptic curves have the same number of points.
(<==) Similarly, if the elliptic curves have the same number of rational points, their Frobenius traces a_{q} are the same, so their characteristic polynomials of Frobenius are both T^{2}  a_{q} T + q^{2}. It follows from HondaTate theory that they have the same number of points.
Since it is known exactly what the possibilities for #E(F_{q}) are in terms of q  this the HasseDeuringWaterhouse theorem  it seems that it is known exactly how many F_{q}rational isogeny classes there are. For instance, when q = p is prime, the theorem asserts that any integer N satisfying the Weil bounds
p+1  2\sqrt{p} < N < p+1 + 2\sqrt{p}
occurs as #E(F_{p}) for some E/F_{p}, so it seems that there are exactly
1 + 2*floor(2\sqrt{p})
F_{p}rational isogeny classes.
For abelian varieties of dimension g, to determine the Frobenius polynomial and hence the F_{q}rational isogeny class, you need to know not just #A(F_{q}) but #A(F_{qi}) for 1 <= i <= g.
The answer for isogeny over the algebraic closure is different  let me know if you're interested in that.

$\begingroup$ Thanks! I'm certainly also interested in the story over $\overline{F}_q$. $\endgroup$ – Rebecca Bellovin Nov 8 '09 at 9:10
For any genus, as long as you have nonismorphic curves, you have nonisomorphic Abelian variety.
For genus 2, you have the HondaTate theorem, giving the classes of Abelian varieties (and there is a recent paper Howe, Maisner, Nart and Ritzenthaler saying which of them are p.p.a.vs)
 Fro genus > 2, I think classification is at the forefront of current research.