Magnet's energy doesn't go anywhere. It always stays $\frac{B^2}{2\mu}$ (in some conditions). So where does the work come from, the energy to pull metal to the magnet?
If you put piece of iron, or better a wire with flowing current $j$, on distance $d$ from source of magnetic field $B$, then there will be force between two: $\vec F=\vec j \times \vec B(d)$ (see Lorentz law). If you hold this wire in place, then there will be potential energy $U=-\int_{0}^{d} Fdx$. If you let this wire go, potential energy will turn into kinetic energy of movement $E=\frac {p^2}{2m}$.
That is how you apply conservation of energy: there is potential energy of B-field, and there is potential energy in system magnet-wire. No wire -- no second part of potential energy.
Now, what does piece of wire with current has to do with real-life piece of iron? See page on ferromagnetism for more, but briefly, there are currents in piece of iron due to electrons' spins, as in any other material. But because in iron currents from many electrons are oriented in same direction, cumulative effect is very strong.
P.s.: I come from/practice experimental physics, so might not care too much about details. That is a flaw. Please comment, so I can refine this answer.