If energy sloshes back and forth between the capacitor and inductor, does this mean that current never goes past the inductor?

Yes, current flows past the inductor, and through the whole circuit. However the faster the rate of change of current through the inductor, the larger the back-emf, which opposes the change in current. Here's what happens in the $LC$ circuit:

• We start with all the energy stored in the capacitor. The capacitor starts to discharge, creating a current. This sets up a magnetic field in the inductor, so energy gets transferred to the inductor. Since the current is changing (initially increasing very fast) the magnetic field is changing and there's a back-emf in the inductor which attempts to decrease the current. Just as the capacitor starts to discharge, the rate of change of current is largest. [Mechanical analogy: this is like the point at which a mass on an extended spring is released. Position ( $x\leftrightarrow q$) is maximum; velocity ( $v\leftrightarrow i$) is zero; acceleration is maximum ( $a\leftrightarrow di/dt$).]
• The current increases, but not beyond a maximum value, due to the opposing back-emf. The maximum current is when the capactor is completely discharged; current is flowing all the way through the circuit at this point in time, including the inductor. At this point, all the energy is in the inductor, stored in the magnetic field. [Mechanical analogy: this is like the point at which the mass is zooming through equilibrium. Position is zero; velocity is maximum; acceleration is zero.]
• Eventually the capacitor is fully charged up with opposite charge sign with respect to the initial condition, and current is zero. [Mechanical analogy: this is like the point at which the mass on the spring is at maximum compression. Position is maximum with opposite sign; velocity is zero; acceleration is maximum.]

• The cycle continutes. Charge, current, and stored energy slosh back and forth.

This animation visualizes the oscillation. See if you can draw the sinusoids for $q(t)$ and $i(t)$ corresponding to this circuit.