Simple answer: 3V/uS is probably enough for a 10V peak (20Vpp) output. The slew rate is all about how fast an amplifier can track musical transients. The problem is, it's rarely the amplifier stage itself that limits the amplifier--although the amplifier itself is often limited for other reasons--but the power supply and things like capacitors in the signal path. Because of all that, it's unlikely that swapping a faster amp--all else equal--will do much of anything.
If you want to do some bogus math...
Here's the steady state case:
Usually, you double the number to achieve a realistic representation of the signal when it starts, so...
For a 10 volt peak ouptut, over a full cycle, the voltage changes from 0 to +10 to 0 to -10 and back to 0, or a total swing of 40V. To push this voltage at the highest audio frequency of 20,000 cycles per second requires moving the voltage...
2 * 4 * 10V * 20,0000/S
or 80V * 20,000 / 1,000,000 uS (microseconds)
or 1.6V/uS
Just for fun, this 10 volts is about 7 Vrms. At 85mA, the power output is 0.6 W. This implies a nomical speaker impedance of 85 ohms.
Do it again with a typical power amp of 50V peak.
2 * 4 * 50 * 20,000 V / 1,000,000 uS
or 8V/uS
Of course, that's all theoretical. The actual slew rate of an amplifier isn't in the op-amp spec. It comes from the total design, particularly the speed of the power supply in delivering a burst of current in the initial microseconds of a musical transient.
There's a lot of other stuff that makes it more black art than anything else. In particular, your amp will never have to put out full voltage at 20kHz--there just isn't that much power at those high frequencies in anything but bogus test signals. If the math above holds at all, it's because 20,000 comes out to 50uS, which just happens to be about how fast a musical transient can move. Coincidence? No, but that math's over my head.
