Determining sampling rate in OFDM demodulation
It is incorrect to say that the sample rate is "much less" than the bandwidth of the signal.  It is true to say that it is a "little less".  Even that statement is only true for part of the demodulator.

Demodulators almost always use a sample rate that is higher than the bandwidth for two reasons: to ensure that all of the signal is captured rather than aliased due to Nyquist frequency limitations, and to do some version of early/late gating when syncing with the symbols.  It is very common to sample at twice the symbol rate because that is very convenient for capturing the symbols and doing early/late gating.

OFDM is different because of the FFTs/IFFTs built into the signal.  There is no advantage for the demodulator to do the FFT at higher than the symbol rate, so they don't.  The roll-off in OFDM is very minimal (an example of typical OFDM roll-off is shown below), though, so the sample rate is not much less than the bandwidth.  In some schemes, like 802.11a, the outer frequency bins are not used which reduces the bandwidth of the signal relative to the sample rate, meaning that the sample rate will be higher than the bandwidth even when taking into consideration the signal roll-off.



Also, before the demodulator does the FFT it has to eliminate as much carrier offset as it can.  This is typically done at a sample rate that is higher than the bandwidth for the same Nyquist reasons as a typical demodulator.

EDIT: I see that I forgot to answer some of your questions.  The sample rate for OFDM is exactly equal to the sample rate that the OFDM transmitter used when it inverse FFT'ed the data.  No, the sample rate has nothing to do with the modulation types used inside the OFDM symbols (e.g. BPSK or QPSK).  No, the sample rate has nothing to do with the FFT length.