What is the maximum "wattage" can be pushed into an ATMega328 ADC pins?
This is actually a special case that is undocumented, and in fact not as simple as you would at first think.
The input port itself is not actually what is in question here. Instead every IO pin on the Arduino, be it set to INPUT or OUTPUT, has a set of ESD protection diodes on them. These look like this:



simulate this circuit &ndash; Schematic created using CircuitLab
The &quot;IO MODULE&quot; when in INPUT mode can pretty much be ignored, since it is a very high impedance. Just like a (roughly) 10MΩ resistor to ground. Very very little current will flow that way.
D2 is also reverse biased so that direction can be ignored.
So the current will want to flow through D1 to the VCC rail (+5V).  The diodes have approximately a 600mV forward voltage (which is where the VCC + 0.5 comes from).
So we can recalculate, given those parameters, what the current flowing through the diode will be to the 5V rail:

Voltage difference = 12V - 5V - 0.5V = 6.5V
Current = V/R = 6.5/330 = 19.7mA

The recommended maximum (from what I can gather) is only 1mA.  You have 20x that.
So you will need at least 20x higher resistances. 10kΩ in the upper half and 6.8kΩ in the bottom half.  That will give you 4.85V on the output at 12V and limit you to 0.65mA in a failure situation.
It also gives you an output impedance of a little over 4kΩ which is below the recommended maximum of 10kΩ for the ADC input.
You can further protect your input by adding a 5V zener diode and small series resistor on the ADC pin so if the voltage goes above 5V it clamps it and protects the pin.



simulate this circuit