Is a bleeder resistor on a small circuit mandatory at the decoupling capacitors?
You have identified your issue correctly.

I couldn't figure out exactly what is the maximum current MSP430 can draw on its P1 pin. I found a parameter named "maximum diode current" in the datasheet, which is 2mA, and it is the best guess I can make. However, it is not that this is the current which will be drawn in practice: once the regulator's input voltage will get below ~4.3V, it is hard to predict the rate of discharge.

You can minimize the discharge time by taking smaller capacitors for regulator's input. Why did you add 470uF in the first place? I see in this datasheet (which is the one you should use according to part number in the schematic) that 100nF should suffice. 

If the natural discharge is still too slow, you can add bleeding resistor as you did. You can even consider adding pull-down resistor in parallel to P1 pin. If the active power consumption is of high importance, there are more power effective techniques of pulling the voltage down.

GENERAL NOTE: 

The usage of bleeding resistors is very common for safety reasons. For example, there are SMPS which utilize huge output capacitors. If you disconnect the load and expose the output pins, these caps can (sometimes) store their charge for minutes. The amount of charge is such that a human touching the outputs can die. In cases like this one, there is common practice to add a bleeding resistor (usually power resistor) in parallel to the output capacitors.