MOSFET fast turn-off not enough
Just because M1 turns off (and M3 isn't on yet) doesn't mean that v(Mid) will go low -- what happens depends on the load current (and load inductance).

However in your circuit, that's not what is causing the shoot through current -- notice that in your 2nd waveform, V(gh,mid) rises slightly at around 6.05 us. This is because V(mid) is falling (by M3 turning on), but M1 (which was off) turns back on again. M1 turns of because of drain-gate capacitance which couples the increasing VDS of M1 to the gate of M1. 

You can improve this by reducing the value of the 12 ohm, and/or slowing down the turn on of M3. Basically in these situations you need to ensure the turn-off 'strength' is stronger than the turn-on capability. 

In more technical terms -- when turning on M3, the dV/dt of v(mid) is limited by the Miller capacitance CDG of M3 to dV/dt = I_gate/Cdg. However, similarly the gate current induced in M1 is also I = Cdg*dV/dt, which is a similar amount. To avoid M1 turning on, you need to have a higher magnitude of gate turn-off current than turn-on current.