There are two notions of supersingular prime: one is relative to a fixed elliptic curve over $\mathbb{Q}$, and one is absolute. For any elliptic curve $E/\mathbb{Q}$, a prime $p$ is supersingular for $E$ if $E$ has good supersingular reduction at $p$. Such primes are known to be asymptotically density zero, but infinite in number (by a theorem of Elkies). Lang hasand Trotter conjectured that the number of supersingular primes less than $N$ limits to a conjecture regarding precise asymptotic behaviorconstant times $\sqrt{N}/\log(N)$ as $N$ gets large.
Edit: Thanks to Emerton for pointing out the connection. I'll try to expand on it a bit. The moduli problem of generalized elliptic curves with $\Gamma_0(p)$-structure has a coarse moduli space that is a smooth irreducible curve away from p, but has mod p fiber given by taking a disjoint union of two copies of $X(1)$ (a genus zero curve) and gluing along supersingular points (this description is more or less in Katz-Mazur, chapter 13). A geometric point describing an elliptic curve with j-invariant $\alpha$$\alpha \in \mathbb{F}_{p^2}$ is glued to thea geometric point on the other irreducible component describing an elliptic curve with j-invariant $\alpha^p$. The Fricke involution switches the components, so the quotient of $X_0(p)$ by this involution is a genus zero curve glued to itself at finitely many supersingular points. The quotient has arithmetic genus zero if and only if all supersingular geometric points are glued to themselves - otherwise, the flat modular deformation to characteristic zero yields a smooth curve of higher genus. In other words, it is necessary and sufficient that all supersingular geometric points have no Frobenius conjugates, i.e., that the j-invariants of theall supersingular curves lie in $\mathbb{F}_p$.
More Edit: I should give a more honest reply to Mariano's question, which was originally raised by Ogg in the mid 1970s (and he famously offered a bottle of Jack Daniels to anyone who could solve it). Half of the question has an answer. If we combine the results of Borcherds's paper Monstrous moonshine and monstrous Lie superalgebras with the results of the paper Modular equations and the genus zero property of moonshine functions by Cummins and Gannon, we get the following fact:
Let G be a finite group acting faithfully on a conformal vertex algebra V by conformal symmetries, and suppose V has central charge 24 and character $\operatorname{Tr}(q^{L_0-1}|V) = j(\tau)-744$. Then for any element $g \in G$, the series $\operatorname{Tr}(gq^{L_0-1}|V)$ is the q-expansion of a modular function that is holomorphic on the upper half plane, invariant under a discrete group $\Gamma \subset PSL_2(\mathbb{R})$ satisfying $\Gamma \supset \Gamma_0(N)$ for some $N$, and generates the function field of the quotient curve $\Gamma \backslash \mathbf{H}$. In particular, the quotient curve is genus zero.
The monster simple group arises in this context because I. Frenkel, Lepowsky, and Meurman constructed a conformal vertex algebra satisfying the above hypotheses, whose group of conformal automorphisms is the monster. The fact given above implies that for each prime p dividing the order of the monster, the quotient $X_0^+(p) = X_0(p)/\langle w_p \rangle$ is genus zero, where $w_p$ is the Fricke involution. In particular, each prime dividing the order of the monster is necessarily supersingular.
I know some people who would like there to be a conceptual (read: non-enumerative) explanation for why all supersingular primes divide the order of the monster. So far, the best I've heard is that the monster is really big, while there aren't that many supersingular primes.