One just take the inverse limit of $\mathbb G_m$ under the inverse system of maps $\mathbb G_m \to \mathbb G_m$ by raising to a natural number power. Or, equivalently, the sequence $\dots \mathbb G_m \to \mathbb G_m \to \mathbb G_m \to \mathbb G_m$ where the $n$th-to-last map is raising to the $n$th power.
The character group of this is the forward limit of the character group of $\mathbb G_m$ along the dual maps. The character group of $\mathbb G_m$ is $\mathbb Z$, and we haver arranged things so the forward limit is $\mathbb Q$.
It is clearly diagonalizable, as it is an inverse limit of diagonlizable (i.e. commutative reductive, I guess) algebraic groups.
It is not too hard to check this is unique.
In general, for any discrete group, you can construct a diagonalizable pro-algebraic group with that character group, by taking an inverse limit over finitely generated subgroups.