For a finite field $\mathbb{F}_p$ ($p$ a prime), is there an asymptotic estimate for the number of ordinary elliptic curves over $\mathbb{F}_p$ up to isogeny?
It is well-known that two ordinary elliptic curves are isogenous if and only if the endomorphism algebras are isomorphic. So if we phrase this in terms of the endomorphism algebras, this boils down to counting the number of imaginary quadratic fields in which $p$ splits into principal prime ideals. Not sure if that is any easier.
Isogeny classes of elliptic curves over finite fields correspond in a 1-1 way to their Frobenius trace $a_p$. By Hasse's theorem, then $|a_p| \le 2\sqrt p$ (so there are $2\lfloor 2\sqrt p\rfloor + 1$ total possible traces.
An elliptic curve is supersingular if and only if $a_p \equiv 0 \pmod p$, and as soon as $p \ge 5$ we have $p>2\sqrt{p}$, so the only way $a_p \equiv 0 \pmod p$ is if $a_p=0$.
Thus there is only one isogeny class of supersingular elliptic curves and $2\lfloor2\sqrt p\rfloor$ ordinary ones.
One explanation for why every frobenius trace occurs in this case is theorem 4.1 of http://www.numdam.org/article/ASENS_1969_4_2_4_521_0.pdf (Waterhouse - Abelian varieties over finite fields) this has a lot more detail on different cases, e.g. $k= \mathbf F_{p^a}$ for $1< a$, letting $q=p^a$ it states:
Theoren $4.1 .-$ The isogeny classes of elliptic curves over k are in one-to-one correspondence with the rational integers $\beta$ having $|\beta| \leq 2 \sqrt{q}$ and satisfying some one of the following conditions:
(1) $(\beta, p)=1$
(2) If $a$ is even $: \beta=\pm 2 \sqrt{q}$
(3) If $a$ is even and $p \neq 1 \bmod 3: \beta=\pm \sqrt{q}$
(4) If $a$ is odd and $p=2$ or $3: \beta=\pm p^{\frac{a+1}{2}}$
(5) If either (i) $a$ is odd or (ii) a is even and $p \neq 1\bmod 4: \beta=\mathrm{o}$
with $a=1$ this covers all $p$ and $\beta$.
