This is a terrible circuit and will not give you anywhere near the 5V@330mA specified.
Simple envelope calculation to prove this: just use Ohm's law with the full 12V across the input 100 ohm resistor. The maximum current which can flow through this is 120mA. More practically, I'd expect you to get only 60mA max at the output assuming you had a perfect linear regulator, and more likely only 40mA max at the output.
Other things I've noted:
- C1 and C2's values are massive. 330mF on the input and 100mF on the output? really? More typical values are in the 100's of microFarads. In theory having large caps doesn't hurt, but ones this large either have very low voltage ratings and/or are much more expensive.
- The two resistors really just shouldn't be there.
- C3 and C2 are redundant. Get rid of one of them. Note that if you remove the 100 ohm resistor you need to remove the zener diode as well!
- Some mobile phones won't charge just because you give them 5V to a USB port. You need to connect the data lines as well. How to do this is beyond the scope of this answer.
- This is really pushing a linear regulator to its limit (probably beyond its limit). In the best case you're going to be dissipating 660mW in the regulator, and at the worst you'll be dissipating 3.1W. That's a lot of heat. You'll certainly need some sort of heat sinking in place.
As far as the absolute minimum voltage ratings requires, C1 must have a voltage rating of at least 7.2V in this circuit and C2 must have a minimum voltage rating of 5V. However, you'd have to intentionally go out of your way to find ones rated this low, and it's a bad idea to run caps so close to their max voltage. It's very common to find ones rated at 16V, 25V, 50V, or even higher voltages for pennies. Use these instead.
I would highly suggest you get some pre-made module (preferably one with a DC-DC buck converter rather than a linear regulator). One example is this module. This doesn't solve the data-line issue (4), but it does solve all of the other issues I listed.