Really? Can't you afford a zener?
On normal inputs there is a diode to Vdd and a diode to Vss. So any input current goes into the power supply through these diodes. So if the microprocessor is consuming LESS than your input current, the current ends up in the power supply where a series-type voltage regulator can't do anything about it, and then you have an excessive power supply voltage.
The two diodes in series give you a PNPN junction structure just like a thyristor, so excessive current through either of the diodes is like the gate current of the thyristor which then latches ON.
5V tolerant inputs are different - I asked what the protection circuit consisted of and got no response (no surprise there!)
I think it must be like a zener, but I don't know for certain, so I would be much happier using my own external zener.
I suppose you could use the input pull-down option to cancel your input current, but as that can vary from 10uA to 100uA the input voltage could end up anywhere, so that's not a great idea either.
If you do use an external zener, don't forget that the external voltage must get below 1V for it to be certain to read a logic zero.
If you know the possible range of your external voltage, then it is best to use a potential divider. That way you can be certain of your logic 1 and logic 0 thresholds. So it's cost you ONE RESISTOR!
The current through the divider will be determined by the pin input capacitance and the speed that you require.
Maybe you need a capacitor in parallel to remove any transients?
If you don't know your external voltage, but just know that it's larger than 5V, then it will require a little more thought; and perhaps more than one extra component!