Just a bit of preliminary theory.
As you probably know, without any flyback diode, be it a rectifier or a Zener, you'll have a (theoretically infinite) kickback voltage from the inductor (valve coil, relay winding or whatever) whenever you try to interrupt its current abruptly. In reality the kickback won't be infinite because the spike will trigger any sort of nasty effects in the circuit it is connected: it will generate electric arcs, it will drive semiconductors in destructive breakdown, it will fry resistors or punch through capacitors dielectric, etc.
All this in the attempt of get rid of the energy stored in the inductor, which is
\$ E_L = \frac 1 2\, L\, I_L^2 \$
where \$I_L\$ is the instantaneous current at the time immediately before the (attempted) switch-off.
Putting a rectifier in parallel with the coil is the standard low-speed countermeasure, as you know. Assuming the diode can stand the inrush current pulse generated by the kickback, it will clamp the voltage across the coil to a safe ~0.7V. Why is it slow? Because at that voltage level (a diode forward drop) and with usual forward resistance values the power dissipated is low, so it takes more time to convert \$E_L\$ into heat.
Using a Zener is faster essentially because it allows the kickback voltage to rise more before clamping it. Of course the Zener voltage must be chosen not to be dangerous for the rest of the circuit. Since the clamp happens at higher voltage, and the breakdown dynamic resistance of a Zener may also be lower, the dissipated power is bigger, hence it takes less time to convert \$E_L\$ into heat.
If you wonder what happens when the clamp action ceases because the current is not enough to keep the Zener (or the clamp diode) in breakdown (conduction), well the answer is that it will probably oscillate, because the energy MUST be converted, since the power source of the coil has been cut-off, and the stored energy depends on the current in the coil. The coil won't "hold the energy" as a capacitor would do, because for that to be possible a current should flow into the coil itself. Therefore the remaining energy will find other ways to get converted: stray capacitance and leakage current of the diodes and parasitic capacitance of the coil itself (for example). It is sort of a non-ideal non-linear tank circuit, which will exhibit damped oscillations until the energy is completely converted into heat.
EDIT
(In response to a comment from @supercat)
Here's some results from a hastily conceived circuit simulation using LTspice showing the damped oscillation that may arise in a situation similar to the one described above.
The transient analysis produces the following plots:
If we zoom in the interesting parts we have:
In the following extremely zoomed-in plot you may notice the estimated frequency of the oscillations (I've enhanced the image to show where LTspice cursors are placed).
As for your #2, there is nothing "hard" about choosing the value of the resistor if you understand what is going on.
You take the desired peak voltage across the resistor and divide it by the operating current of the coil. That's it- good old George Simon Ohm's Law. The current immediately after the switch turns 'off' is the same (or a bit less) than it was before, so the voltage the switching device will see is the power supply voltage plus a diode drop plus the operating current times the resistor.
A bipolar TVS could be used in place of the zener + diode if the voltage you wish to clamp at is substantially more than double the power supply voltage. TVS devices are designed to clamp voltage under high current conditions so they're made to withstand massive surges in power (often in the 1kW range). A normal coil doesn't have that kind of surge, so it's a bit of a waste. If you put (say) a 6V bipolar TVS across a 12V coil intending to clamp to 18V, you will have a bad day, since the TVS will conduct in parallel with the coil and probably destroy the driver.
You could use a unipolar 18V TVS across the transistor though.
Best Answer
You must model the solenoid valve as a resistor (33\$\Omega\$ in your case) in series with an inductor. If you like you can include a bit of parallel capacitance, but the resistance is required.
Otherwise the simulated current after a long time will be limited only by the MOSFET (maybe to something like 30A) and the voltage will approach zero.
For the operation time of the valve, you can consult the data sheet. Normally they will specify it without a diode, so whatever time you get with the zener will be a bit longer. You can guess that the difference in the current through the coil will give you an idea of the operation time, but there is no guarantee that even if the coil current drops to 10% or 5% of normal that the mechanical parts will have moved in that time. Some valves are fairly complex ('pilot valves', for example that operate indirectly), and the inductance of electromechanical systems often varies significantly during operation.
Edit: Below is a PSPICE simulation I did with similar parameters to yours (I used a 1N4740 Zener and a PHD23NQ10T,118 MOSFET to save me time). Green trace is the gate drive signal (before the resistor), red is the MOSFET drain. As you can see the MOSFET drain rises to a peak voltage of around 24V.
In the below close-up, you can see where the bulk of the energy is dissipated and the ringing afterward. The violet trace is the coil current (scaled so you can see it on the same graph).