Using two polarised capacitors back to back to make a non-polarised capacitor of half the value is an old tecnique, but not one I like much personally. Quite why you want the first DC blocking cap is a mystery as firstly mains doesn't have any DC, and secondly a bridge rectifier doesn't mind some DC on the input. It does add about 640 ohms in series resistance, which is probably it's purpose.

The reservoir capacitor on the output of the bridge can be polarised, as the bridge generates a positive and negative output. The reservoir capacitor needs a 400V rating, and I would use just a single polarised capacitor.

As you have calculated, 72 LEDs drop about 273V. Assuming a maximum AC RMS input voltage of 264V, the reservoir may get up to 373V, so you are looking to drop 100V through the series resistance. Calculating the LED current is quite tricky for the circuit shown, but it looks like it's about 100mA if you include the input capacitor. This means that each 470 ohm dropper resistor dissipates nearly 5W. If you manage to adjust the components to give 350mA, then the dropper resistor will dissipate nearly 50W!

Note that a mains transients will rip through the series capacitor and might blow all your LEDs.

By the time you've added tarnsient suppression, heatsinks and isolation for a control transistor, it might be cheaper to use a proper isolated PSU.