Trying to hold up the regulated 5 V line with a capacitor is the wrong way to address this problem. The USB power will sag somewhat, and a very large capacitor will be needed to not let it sag too much.
Do the math. USB devices can draw up to 500 mA. Let's say you don't want the power voltage to sag more than 500 mV, and that you want to be able to ride out at least a 2 second cranking event.
(500 mA)(2 s)/(500 mV) = 2 F
Note that the power voltage will always sag, and that the 500 mV spec was arbitrary since we don't know how close to the valid USB power voltage lower limit you are already at.
A better answer is to hold up the input voltage. Put a Schottky diode in series with the 12 V, then a capacitor to ground before the USB power supply input. These small USB power supplies are switchers, and will work with lower input voltage. You don't know how low, but that is something you can measure while loading the output with whatever current your device needs.
Using the same parameters as above, you want to ride out a 2 s interruption in 12 V input while drawing 500 mA from the USB. That comes out to 5 J of energy output with no corresponding input. Let's say the power supply is 85% efficient, so you need to feed it 5.9 J. For sake of example, let's say the power supply still produces 5 V out with 8 V in. Doing the math yields 150 mF.
That's also a large capacitor. Either way the capacitor has to store enough energy to run the load during the time the 12 V power is off. However, the advantage is that the USB power voltage never sags at all. The device is guaranteed to continue running.
This also points out why batteries are usually used for devices that plug into car power but need to keep running during starting. It's just too much energy for a reasonable capacitor to store. Note that the examples above used only 2 seconds for the 12 V off time. That's probably good enough for most cases, but sometimes it will take longer, and your device will get powered off.