One way to think of $\mathbb Q$ is the direct limit over positive integers $n$ of $\frac{1}{n} \mathbb Z$. Thus giving a character of $\mathbb Q$ is the same
as giving an element in the projective limit of the character groups of
$\frac{1}{n}\mathbb Z$. In particular, if we restrict to unitary characters,
we find that $\mathbb Q^{\vee}$ is the projective limit of circle groups $S^1$ under the $n$th power maps. This object is (I think) called a solenoid; to number theorists it is better known as the adele class group $\mathbb A/\mathbb Q$. (Here and throughout I am using the discrete topology; if one instead considers the
induced topology from $\mathbb R$, then, as Robin explains, one just gets
characters of $\mathbb R$.)
The exact sequence $0 \to \hat{\mathbb Z} \to \mathbb Q^{\vee} \to S^1 \to 0$
in Pete's answer arises from the map taking the solenoid to the base $S^1$; the fibres of this map are copies of $\hat{\mathbb Z}$.
If we wanted not necessarily unitary characters, we would instead get
the projective limit of copies of $\mathbb C^{\times}$ under the $n$th power maps. Since $\mathbb C^{\times} = \mathbb R_{> 0} \times S^1$, and since
$\mathbb R_{> 0}$ is uniquely divisible, this projective limit is simply
$\mathbb R_{> 0}$ times the solenoid.
On a slightly tangential note, let me remark that
the relationship with the adeles is important (e.g. it is the first step in Tate's thesis):
Since the adeles are the (restricted) product of $\mathbb R$ and each $\mathbb Q_p$, and since these are all self-dual, it is easy to see that $\mathbb A$ is self-dual.
One then has the exact sequence
$$0 \to \mathbb Q \to \mathbb A \to \mathbb A/\mathbb Q \to 0$$
which is again self-dual (the duality swaps $\mathbb Q$ and the solenoid
$\mathbb A/\mathbb Q$).
One should compare this with the exact sequence
$$0 \to \mathbb Z \to \mathbb R \to \mathbb R/\mathbb Z = S^1 \to 0.$$
This is again self-dual ($\mathbb R$ is self-dual,
and duality swaps the integers and the circle).
This brings out the important intuition that the adeles are to $\mathbb Q$ as
$\mathbb R$ is to $\mathbb Z$.