The point is that, if you dump 4500 watts into a speaker and only produce 108 dB SPL, that speaker is horribly inefficient. By comparison, the power amps in my Event ASP8's have a combined output of 280 watts, and I've measured 111 dB from 1 meter away.
Another point: that power outlet in your wall—the one that the Phantom needs to plug into to make sound—can only deliver 1800 watts. If the Phantom puts out 4500 watts, where does the extra 2700 watts come from?
AC mains supply has basically nothing to do with power output.
An amplifier is defined as a device that amplifies both voltage and current in a way that the power out to the speaker divided by the audio power into the amp is greater than 1. Since the audio power of the signal into an amp is in the range of a couple of mW, and the output power is in watts, this condition is pretty easily met.
Second, the amplifier has a power supply that operates at a different voltage(s) than the AC supply, therefore currents do not translate 1:1. Also, the amplifier's power supply performs an averaging operation on the AC supply, and most are chosen so that the averaging function and the duty cycle of the audio work hand in hand (to a reasonable degree).
Third, without knowing the duty cycle of the signal being amplified, you can't calculate average audio power... which is necessary for the AC supply calculation.
The math is simple for an engineer who does this all the time, but it's probably not terribly intuitive the first time through, maybe the second or third as well. We have to perform these calculations when designing amps and power supplies, when developing the UL/CB Scheme test protocols, and for some of us who work with pre-packaged SMPD/Class D modules it's even more important to understand these dynamics and how control of these parameters are passed back and forth between the SMPS and amplifier as there is additional processing happening in some of these cases.
The easiest way to do this is to base everything on the power equations rather than jumping between voltage and current all the time. Below are the standard calculations (for a class D SMPS amp) used for this purpose:
AC power source = 120V x 16A = 1920 Watts
AVERAGE audio power out = Audio Power x duty cycle = P x .125 (using the 1/8-rated power per UL/CB Scheme)
AAP x amplifier efficiency = power draw on DC power supply = AAP x .85 (using a typical 85% efficient amp)
AVERAGE AC power = power draw on DC power supply x SMPS efficiency = AACP x .9 (using a typical 90% efficient SMPS)
So solving the entire string of equations:
(1920W x .9 x .85)/0.125 = 11750Watts
Back to the 4500W
Let's use an average power supply efficiency of 90% and an average amplifier efficiency of 90% (pretty typical of this amp, though efficiency is not a single number and through some unique control integration between the power supply and the amplifier, in many areas as the amp efficiency rises the power supply efficiency falls and vis-versa yielding an almost flat 81% efficiency (which happens to be the same as 100 x .9 x .9 = 81%). Let's now increase the duty cycle from 1/8 to 1/3 which is very demanding. The equation solves as: (1920W x .9 x .9) / 0.333 = 4700W of audio power.
I could go on, but I don't know how much EE knowledge you have. No offense meant at all, but judging from your initial inquiry, I'll assume it's not very high (or at least, not in audio amplification). To put very simply - the power supply voltage is not anywhere near the same as the mains voltage, hence, your standard Ohm's law cannot be applied without a tad more investigation.
I also don't doubt that they're inefficient. They're targeted for home audio reproduction, not live music production. Two very different design goals.