Statements from ATmega320 datasheet
The ADC is optimized for analog signals with an output impedance of approximately 10 kΩ or less.
Voltage on any Pin except RESET with respect to Ground ....... -0.5V to VCC+0.5V
So the protection resistor for analog inputs should not be higher then 10K. Just imagine this very very very bad situation. You have an analog input that gives you a max. of 12V+ and 12V-. You could handle the over-voltage using a voltage divider. The problem is still the negative voltage.
Using a voltage divider the analog input might still see minus -5V. Using a 10K protection resistor will reduce the current thru the protection diode to max. 0.5 mA. The voltage drop at the protection diodes is 0.5V. It is not specified but you can deduct this from the max. input/output voltages. (Schottky diodes?)
5V*0.4mA=2mW. This sounds like almost nothing. But do not forget that the protection diodes are extremely tiny spots on a chip with 1000s of transistors. If the chip is busy it might be quite warm already. Think of a local hotspot.
The ATmega328 datasheet has a circuit diagram of the protection unit (page 313 in version Rev. 8161D). It does NOT give you any information how many milliamperes the diodes can handle.
Protecting from a positive over-voltage is quite easy. Maybe just using a voltage divider or a zener diode with a resistor in front. Protection from negative voltages is not that easy. A regular diode in front of the analog input will will not help when you use a diode with a forwarding voltage higher then the protection diodes. So I do miss some information in the datasheet.
However, I think it can handle at least 1mA. Just a guess. So a 10K resistor will probably do the job.