For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire. For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.
If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field. In this limit, you will obtain a uniform electric field through all space.
Now, put your conductor in place along the axis between the voltage sources--a current will flow. In the DC case, this gives rise to the magnetic field outside of the wire. As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance. Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor. With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current. Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.
For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.
For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential. The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.
I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit. Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.
The work comes from the battery that is driving the current through the wire.
Even if the wire were stationary, the battery would be supplying work at a rate $I^{2}R$. But with the wire moving, the battery would need to be supplying extra work at a rate $\mathscr{E}I$ in order to overcome the emf generated by the moving wire.
Now, $\mathscr{E}$ is equal to the rate at which the wire cuts magnetic flux so $\mathscr{E}=BLv$ (in which $v=\frac{d}{t}$), so the extra rate of doing work has to be $\mathscr{E} I=BLvI=BLdI / t $. And this is equal to the rate of mechanical work done on the wire!
Best Answer
The claim, "the magnetic force does no work," while technically true, is misleading in so many situations and actually helpful in so few that we probably shouldn't put as much emphasis on it as many textbooks do. The magnetic field stores energy, and a change of the configuration of a system (like moving a current) can convert that stored energy into other forms, so the system will certainly behave as if the magnetic field is doing work, but careful analysis will always show that it's not actually the magnetic force that's doing the work associated with the change of magnetic energy. Usually it's an electric force.
In your example, I think the subtlety lies in assuming that $I\vec{\ell}$ for the current $ = q \vec{v}$ for individual charges. The charges will tend to change trajectory, i.e. $\vec{v}$ changes, due to the magnetic field, but $\vec{\ell}$ maintains its direction. Why? Because the wire (magically? We never really explain it very well in class) constrains the charges to move along its length in spite of their lateral movement produced by the magnetic field. As others have said above, the wire exerts a force on the charges to keep them in a straight line, and so the charges exert a reaction on the wire which moves it.
So technically, the magnetic force isn't doing any work on the current; it's just redirecting the force that would otherwise be pushing the current down the wire so that that force has a component pointing perpendicular to the wire instead.
Now for the bonus question: a magnetic dipole, e.g. a current loop or a small refrigerator magnet, will experience a force in a nonuniform magnetic field (though not in a uniform one). The dipole will then accelerate if this is the only force acting on it, so work is being done on it. If it's not the magnetic force doing the work, then what is it? (Really, as in this case, it gets silly sometimes to insist the magnetic force isn't doing work. For many intents and purposes, you might as well treat it as though it is.)