One problem with your present circuit is that by putting the N-channel MOSFET on the "high side" of the tank circuit (between the power supply and the tank), it is dissipating far more power than necessary. You're essentially forcing the drain-source voltage to be equal to the gate threshold voltage (about 4V, based on your waveforms), rather than allowing it to be as low as possible.
One obvious solution would be to put the MOSFET on the low side of the circuit, between the tank and ground. Or you could stay with a high-side switch, but make it a P-channel MOSFET instead (which will require an inverted drive signal).
Either way, the MOSFET will stop conducting as soon as the gate pulse ends. However, this means that you may see some very high (or low) voltages at the drain of the MOSFET because of the inductive kick of the coil. You will want to add something to the circuit that limits the voltage to whatever the MOSFET can tolerate — perhaps a large-value zener diode.
Just to put some numbers to this, and assuming zero losses, the peak current in the coil will be
$$I_{peak} = \frac{V}{L}\cdot t_{ON}$$
And the peak voltage after the MOSFET cuts off will be
$$V_{peak} = I_{peak} \sqrt{\frac{L}{C}}$$
Which means that you can control Vpeak by either limiting the on time of the gate drive signal, or controlling the ratio of L to C, or a combination of both.
Using some numbers pulled from your scope traces, it looks like if your capacitor is 5 µF, your coil must be about 2.5 mH. Also, your tON looks to be about 1.6 ms.
Therefore, Ipeak is going to be about 7.68 A (!)
Vpeak will be about 172 V.
Without a proper load, there is nothingc to pull the drain high when you turn the MOSFET off, except the 10 megohms of your scope probe. That and the FET output capacitance give rise to the slow decay you see.
Make the same measurement with a resistor ( 1 kilohm or lower) in parallel with the probe, and observe the difference..
Best Answer
Let's go over the main specs of the MOSFET...
Drain-Source Voltage (Vdss)=60v. You will be subjecting it to 5v, so you're good!
Gate-Source Voltage (Vgss)=20v. You're at 5v, so good.
Drain Current (Id)=300 mA continuous. You are at 100 mA (20 mA for each of the 5 LED's). Good!
Gate Threshold Voltage (Vgs(th))=1.0-2.5v. You'll be driving the gate with either 0.0v or 5.0v, so here again you are good.
I ignored some of the more esoteric specs, which could come into play if you were driving an inductive load like a motor or relay, or needing to have fast switching or high-frequency switching. But for turning some LEDs on or off, this MOSFET will be just fine.