The cheapest way is a diode in series with the 3.7 V LiPO. That will drop about 0.6 V (the MSP430 will probably not consume much power). A resistor's voltage drop will be less constant. The MSP430 will operate at voltages as low as 1.8 V, and then the camera will have given up already.
To simulate a button-push you can control a transistor which you mount parallel to the button, from the MSP430.
The diode will give you a constant voltage drop, so if the battery's voltage sags to 3.5 V the MSP430 will get 2.9 V instead of 3.1 V. It doesn't matter much for its operation, like I said it will keep working at a voltage as low as 1.8 V. (At work I tested how low it would go, and it still worked at 1.3 V, but that's no longer guaranteed.)
If you want to be a Good Boy you would use a voltage regulator, in this case an LDO, for Low Drop-Out. Many voltage regulators are three pin devices: 1 pin input, 1 output, and a common ground, that's the negative pole of your voltage. Standard regulators need a few volt more input than output, so for 3 V out you may need 5 V in, more than the battery can supply. An LDO will work with less voltage difference. For instance a 2.5 V LDO may need only 2.7 V input. The regulator will keep the output at 2.5 V as long as the input is higher than the 2.7 V. You'll also need a couple of capacitors, one for the input, one for the output.
Oli's MCP1700 is a good choice. It has a very low ground current, together with the MSP430 you should be able to stay below 10 µA, so you'll barely discharge the battery. 44 cent at Digikey (Farnell is 55 cent).
A 1N4148 diode costs 10 cent.