Since the TPIC6B595N registers can only sink current, they are connected to PNP transistors on the positive side to source the current.
That's fine, but it's a bit of a waste. TPIC6B595N is notable in that it's a power shift register. A quick read of the datasheet suggests that each output can sink 150mA continuously. This is way more than the ubiquitous 74HC595, which is way cheaper. I'd go ahead and use TPIC6B595N on the bottom, and use the much cheaper 74HC595 driving some power MOSFETs on the top.
This setup is explained in more detail: Explain the use of NPN and pMOSFET in this 8x8x8 LED cube. You will need to select a P-channel MOSFET that can handle at least all the current required when all the LEDs in that section are on. Your conclusion of this being \$7.6A\$ appears sound.
the datasheet suggests 150 uA for each register so the current needed for the ICs may be insignificant.
I don't want the LEDs to burn if only a few of them are on so how do I prevent them from drawing more than 20mA of current?
You need a current-limiting resistor in series with each LED, or some other current-limiting component. Keep in mind you can't share resistors between parallel LEDs.
How would I go about distributing the power over various components if they have different voltage and current ratings?
Your power supply will pump whatever current is required to maintain 5V, so your life is simplest if everything can operate with 5V. Shift registers for 5V operation are probably the most common kind. Select your LED resistors for a 5V supply and you are all set. I don't think you have any other devices which do not operate from 5V.
One final consideration: parts of this circuit will be carrying some significant currents. Be sure to check out a trace width of wire gauge calculator (Google has hundreds) against the high-current traces. If your traces or wire aren't fat enough for the high-current paths, you will lose a significant fraction of the supply voltage and energy in the wire, and in the worst case, start a fire.
I would hold out for a Z-series Vishay metal foil resistor, which are available in 100\$\Omega\$, nominal 0.2ppm/degree C.
Mount with the recommended (see mfr application data) lead length to the Y connections, heat sink during soldering and fill your box with dielectric foam (eg. Urethane) to keep air currents off. If you are paranoid, use low thermal EMF solder, but try to keep everything as symmetrical as possible in any case. You can/should get it calibrated by a lab if it's important.
A resistor will reach a thermal equilibrium when the dissipated energy equals the energy drained to the environment. To drain heat to the environment the resistor's temperature has to be higher than the environment's; the higher the temperature difference the more heat will flow. So the resistor can dissipate more power at low environment temperatures. Rated power may be specified at 25°C and derate at higher temperatures. These thick film chip resistors derate only from 70°C, as the following graph shows:
So a 100mW 0603 can still dissipate these 100mW at 70°C, but shouldn't dissipate more than 50mW at 100°C environment temperature.
The datasheet doesn't give suggestions towards copper layout (land patterns and trace widths) which will influence conducted heat. (Convection will be low for SMDs, and radiation almost zero; the temperature is too low for that.) It may be tempting to have lots of copper connecting to a pad, but make sure that this doesn't cause trouble in soldering.