What you have to keep in mind is that the impedance of a coil varies strongly with the frequency.
You measured the inductance of the coil to be 440 microhenries.
440 microhenries has an impedance of 4 ohms only at about 1500 Hz. If you get much above that, the impedance goes up. If you get much below that, the impedance drops.
You need to keep the frequency in the vicinity of 1000 to 2000 Hz with that coil.
You mentioned in another post that you are using an SMSL SA 98E amplifier. It uses a TDA7498E amplifier IC.
That amplifier uses a class D bridge output. Clipping the ground wire of the scope to one of the output wires may cause a short in one side of the bridge. That would cause your coil to be exposed to DC - and your coil is basically a short circuit at DC.
A large capacitor in series with the coil would help in that case.
The best thing to do would be to try it without connecting the scope for now. If the coil stays cool when the scope is disconnected, then there's ways to make the measurements.
If the coil alone still gets hot, you might put a large capacitor in series with the coil. I'd try about 50 microfarads. That'll give you about 3 ohms impedance at 1000 Hz added to the impedance of the coil. That should prevent overheating in the coil from any DC that may be present.
You've confirmed it was a problem with the oscilloscope.
Connect a 100 nanofarad capacitor to the probe and another to the ground lead of the probe (that's two capacitors.)
Connect the free ends of the capacitors to the coil.
Your scope should show the driving signal with the correct shape and voltage.
For a start, you can estimate the power from the voltage and the impedance of the coil.
You measured it atinductance of the coil and got 440 microhenries. Use that and the driving frequency to calculate the impedance (\$Z = 2 \pi fL\$ where Z is in ohms, f is in hertz, and L is in henries.)
Measure the voltage with the scope. Calculate RMS voltage from the peak to peak voltage, or use the RMS function if your scope has it.
Effective power is \$P = \frac{V_{RMS}^2}{Z} \$, in watts, where V is in volts and Z is in ohms.
That won't be perfectly accurate, but it'll give you numbers that are at least somewhat realistic.
That'll be a bit inconvenient because of the math - if you change the frequency you have to start all over again with calculating the impedance and then the power.