If they are the same kind of switches I have, leg 2 will not be connected to anything while the switch is in the middle position. That means the analog input is open, and in the hardware world (as opposed to the Axoloti patcher) an open input does not automatically become zero. Instead it is "floating" or undefined. Typically an open analog input reads the same voltage as the pin next to it, or any random voltage from static electricity on the wire or even the signal of the nearest AM radio transmitter.
So you should make sure that the analog input is never totally unconnected. In this case, a pull-down resistor would help, i.e. a resistor between the analog input and ground. Of course that resistor would then in turn influence your VDD/2 (when the switch is in that position), so it needs to be relatively high.
A better solution: connect leg 1 to VDD, leg 2 to analog in and leg 3 to GND, then add one resistor between leg 1 and 2 and another resistor between leg 2 and 3. That way the analog input is always connected to a well-defined voltage and each switch has its own voltage divider, so they don't influence each other. As to the resistance value, I would use 10 kOhms (because I have those lying around already), but anything between 1kOhm and 100kOhm should work. Lower resistance values draw more power, higher ones give a slightly unstable signal.