Very interesting question. Lets consider an ideal case of a long spring with a high density of turns in the spring and no friction or air resistance on the mass (so no damping from air resistance or transfer of energy to internal modes of energy). We will use a cylindrical geometry $(r,\phi,z)$. Assume the free spring constant is $k_{0}$.
So let us consider a vertical spring with a mass on it. Also, we are running current $I$ through the spring. The spring has a high density of turns $n$, so lets assume each turn as a circular current loop. The spring is long too, so the magnetic field $\boldsymbol B = B \hat{z}$ generated by the current in the spring is perpendicular the the loops. Then the total lorentz force between the adjacent current loops $$F=\int I (d\boldsymbol l \times \boldsymbol B)$$
is zero, since the forces on opposite sides of the loop cancel (for instance, the force on the top of the loop points up, and the force on the bottom points down). In this idealization, there is actually no magnetic force between adjacent spring loops. In this case, the spring will oscillate up and down forever if the mass is displaced from its equilibrium point.
Ok, great, but in real life the magnetic field is not so uniform, i.e. $\boldsymbol B = B_{z} \hat{z} + B_{r} \hat {r}$ near the ends of the spring. In this case, yes, there is a Lorentz force on the ends of the spring, compressing it. You can imagine this spring as a stiffer, smaller spring. In lieu of an in depth analysis, my guess is we can approximate this spring as a new spring with a different equilibrium point and higher spring constant $k_{1}$, that, under small oscillations, will still behave like a normal spring.
I would suggest starting with a simple force balance $m \frac{dv} {dt} = -kx + F_{mag}$ and see what equations of motion you can get. A more interesting endeavor might be to formulate a Lagrangian for this Newtonian/Electrodynamic system.
