Very confused regarding parallel resistors limiting current
The only way a parallel resistor limits voltage is by loading down the source. It makes the power supply need to provide so much extra current that it can no longer maintain its output voltage. It's like reducing the speed of your car by filling it with steel plates so it is weighed down.

The mechanism by which this occurs is due to the internal resistance in series with the output of the source. If it is too high then too much voltage drops across this internal resistance at higher output currents and therefore less voltage appears at the output of the source.







simulate this circuit &ndash; Schematic created using CircuitLab

In other words, don't expect or try to use a parallel resistor to limit output voltage of a power supply.

BUT, we are not talking about a power supply here. We are talking about a piezo element which is not acting as a power supply so much as a signal source (much lower power than a power supply so efficiency and heat are not real concerns). So the megohm resistor is probably there to load down the piezo element in a similar way to protect the ADC. 

In your case, the piezo would the V1 and R1, while the load would be your ADC (and the 1 megohm resistor if you include one).

Personally, I would use a TVS diode or something similar to clamp the voltage instead of a resistor, but a resistor is much cheaper. It's also possible that the piezo provides way more signal than is actually required so loading it down is a cheap and dirty way to reduce it.

EDIT: As per Elliot's comment if it is indeed true that a piezo is best modelled as a current source rather than a voltage source, then another way that you can look at it is that the 1Mohm resistor is used to convert the current produced into a voltage that the ADC can measure. Without the 1Megaohm resistor, the lone input impedance of the ADC is so high that if the piezo was acting as a current source, the requires voltage to develop the desired current across said impedance would be too high which would damage the ADC.