Let me flesh out my comments. First, why there are infinitely many three-dimensional simple Lie algebras over $\mathbb Q$. One of the major steps in proving class field theory is to prove that we have an exact sequence of Brauer groups
$$0\rightarrow {\rm Br}(\mathbb Q)\rightarrow\oplus_{p\le\infty}{\rm Br}(\mathbb Q_p)\rightarrow \mathbb Q/\mathbb Z\rightarrow 0$$
where for $p<\infty$, ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$, and of course ${\rm Br}(\mathbb R)\simeq \mathbb Z/(2)$. This says that given a finite set $S$ of primes (regarding $\infty$ as a prime) and division algebras $D_p$ over $\mathbb Q_p$ for all $p\in S$, if the corresponding invariants (in ${\rm Br}(\mathbb Q_p)\simeq\mathbb Q/\mathbb Z$) sum to zero in $\mathbb Q/\mathbb Z$, then there exists a division algebra $D$ over $\mathbb Q$ giving rise to the $D_p$ in the sense that $D\otimes_\mathbb Q\mathbb Q_p\simeq M_{n_p}(D_p)$ (we need to use matrices over a division algebra because the dimensions of $D$ and $D_p$ don't have to match).

Now, for quaternion algebras over $\mathbb Q$, the situation is simpler because (1) a central simple algebra of dimension 4 over $\mathbb Q_p$ is either a quaternion division algebra or $M_2(\mathbb Q_p)$ and (2) the hard algebraic number theory can be done "by hand" (it basically follows from quadratic reciprocity). Note this implies that the invariants for a central simple algebra of dimension 4 is either $0$ or $1/2$, hence the need for an even set of primes to get a division algebra over $\mathbb Q$.

So we have infinitely many quaternion algebras over $\mathbb Q$, all of which split over $\mathbb C$ to be isomorphic to $M_2(\mathbb C)$. To get three-dimensional Lie algebras out of this, you restrict to elements with "reduced trace" equal to zero.

Second, we wonder if we have found all three-dimensional simple Lie algebras over $\mathbb Q$. This is somewhat outside of my comfort zone. We switch to talking about simple algebraic groups of dimension three. We are interested in classifying forms of $SL_2(\mathbb Q)$. These are classified by the (non-abelian) Galois cohomology group $H^1(G(\bar{\mathbb Q}/\mathbb Q),{\rm Aut}_\bar{\mathbb Q}(SL_2(\mathbb Q))$. These can further be split into two classes, inner forms and outer forms, depending on whether the corresponding automorphism is inner or outer. Inner forms correspond to quaternion algebras. Outer forms correspond to certain unitary groups. This is from Platonov and Rapinchuk's "Algebraic Groups and Number Theory", section 2.3.4 (propositions 2.17 and 2.18). So it seems like we might be missing a bit, but the simple nature of our situation may mean that the outer forms are isomorphic to the inner forms, like $SU(1,1)\simeq SL_2(\mathbb R)$ (I have no clue).

As Vladimir notes in the comments to this answer, I am assuming that if two quaternion algebras are non-isomorphic, then their Lie algebras are non-isomorphic. This is a legitimate worry, as if $K$ is a quadratic extension of $\mathbb Q$, then ${\rm Lie}(K)\simeq {\rm Lie}(\mathbb Q^2)$. What happens for quaternion algebras? First, note that for a field $k$ and a quaternion division algebra $D$ over $k$, ${\rm Lie}\big(M_2(k)\big)$ is not isomorphic to ${\rm Lie}(D)$, since, for example, ${\rm Lie}\big(M_2(k)\big)$ has a solvable three-dimensional subalgebra and ${\rm Lie}(D)$ does not (seeing this by explicitly calculating the brackets of basis elements). Second, for two quaternion algebras $D_1$ and $D_2$, if a quadratic extension $K$ splits $D_1$ but not $D_2$, then $D_1\otimes_\mathbb Q K\simeq M_2(K)$, but $D_2\otimes_\mathbb Q K$ remains a division algebra. Since ${\rm Lie}(D_1\otimes_\mathbb Q K)\not\simeq {\rm Lie}(D_2\otimes_\mathbb Q K)$, we have ${\rm Lie}(D_1)\not\simeq {\rm Lie}(D_2)$. Finally, since quaternion algebras over $\mathbb Q$ are determined by the primes where they split, we can always find a quadratic extension $K$ so that exactly one of the $D_i\otimes_\mathbb Q K$ splits.

This argument only partly extends to general division algebras, since there are non-isomorphic division algebras with the same splitting field.

After I typed the above, I happened to see that Chapter X of Jacobson's "Lie Algebras" is devoted to classifying simple Lie algebras over arbitrary fields, which he also does in this paper. In particular, he proves that for central simple algebras $A$ and $B$ over a field $k$,

An isomorphism between ${\rm Lie}(A)$ and ${\rm Lie}(B)$ extends uniquely to either an isomorphism or the negative of an anti-isomorphism between $A$ and $B$. If they are quaternion algebras, it is always an isomorphism.

even(finite) set of primes". Also, I think the simplest way to see this is that there is a unique quaternion division algebra over $\mathbb Q_p$, then use quadratic reciprocity to show that you can stitch an even number of them together to form a nontrivial quaternion algebra over $\mathbb Q$. Then the fact that quaternion algebras arising from different sets of primes are not isomorphic is obvious. – B R Oct 18 '11 at 22:50