Class D amplifier - equation for peak current?
Sadly, there is no such simple relationship. Speakers and the amplifier (and the filters) form a resonant system, and that can introduce much higher voltages into the system than supplied by the power supply.

However, it doesn't really matter:

The peak power consumption isn't something that you need to supply, but a design factor that you can choose. It's up to you, within the limits of not damaging your speakers or the switching transistors, to decide how much peak current your amplifier system needs to be able to draw.

For example, you might decide that you want your power supply to deliver enough power for the max RMS power consumption of the amplifier (that's not a "must"! You can built an amplifier that could draw more power than the supply can offer, and then use it at less-than-maximum-volume). That's easy, you just use \$P=U\cdot I\$ and solve for \$I\$, and get a \$U\$ voltage power supply that can at least supply \$I\$ current continously.

Then you decide that at that (undubitably loud) setting, you still want to be able to, for a short term of \$t=0.5\,\text{s}\$, supply more power, say four times as much. So, that'd be a current factor of four. You decide that you need to supply half of that by having half of the appropriate charge (\$I\cdot t = Q\$) in a capacitor, and overdimensioning your power supply by a factor of 2. (just rough numbers; none of this takes efficiency or the exponential discharge curve of a capacitor into account)

In any case, the speaker impedance doesn't come into play here, anyways: your current goes through the switching transistors of your class-D amplifier, so your amplifier manufacturer is in charge of definining a maximum power and/or current, anyway.