So I plan on using the VS1053 Codec connected to a class D Amp (2 x 2.8W channel) with two 3W 4 Ohm speakers. If I supply 5V to this setup, what would the current draw be at full use?
According to the docs the efficiency is reasonably high, I believe 89% for 8Ω is mentioned.
If you want to be able to drive the speakers at full power continuously:
Total power [Watt] = number of channels × power per channel
P = 2 × 2.8 = 5.6 W
Supply current drawn for the speakers alone would be:
Current [Ampère] = Power [Watt] / Voltage[Volt]
I = P / V = 5.6 / 5 = 1.12 A
Accounting for efficiency, you are looking at 1.12 / 0.89 = 1.26A. This is with 8Ω load, efficiency with a 4Ω load will be worse.
The datasheet for the amp states another 5.5 mA supply current.
The datasheet for the codec lists power draw of up to 60 mA.
Total current draw would be at least 1.35A at full load.
However there are quite a few assumptions here and you really want some headroom for unforeseen things. Maybe you want to add a indicator LED? I did not account for an SD card or another microcontroller to control things. In practice you mustn't aim for a power supply that can't do at least 1.5A and I'd probably pick a 2A power supply as it is good practice to not push a power supply to its spec'd limits.
Sorry for the late reply. You cannot simply develiver full power (2.8W) into 8 Ohm speaker by using 5V power supply. So the answer given by jippie is wrong, the assumption is wrong. I suggest you to read this answer by jonk "Max Voltage for Speaker of Given Power Rating"
So you use class D amp with BTL connection, the maximum power you can deliver to 8 Ohm speaker is (5/1.41)^2/8 = 1.57W, of course you should also count loss on amp. You can also see the similar result from amp datasheet:
1.7 W/Ch Into 8 Ω at 5 V (10% THD+N)
2.8 W/Ch Into 4 Ω at 5 V (10% THD+N)
So you can only develiver 2.8W full power to 4 Ohm speaker at 5V.
So I will use this data from datasheet to be used as "full use" to calculate current draw:
THD+N = 10%, VDD = 5 V, RL = 8 Ω 1.72W
For efficiency at 90%, the current draw by speaker amp Iamp = 1.72/0.9/5 = 382mA