An accuracy of +/- 5°C accuracy at 280°C amounts to about 1.8% error. This results in an effective resolution of 6 bits (assuming the full measurement range would end at about 280°C). 10 bits of resolution would result in about +/- 0.28°C accuracy, and 8 bits in about +/- 1°C. So you don't need to worry here (Even when you are not using the full range of the ADC input).
The easiest solution for your overflow problem could be to use Avcc as reference voltage (but then it should be noise-free, precise and stable enough). This reduces your resolution (by half when compared to the internal reference, because it doubles the measurement range), but you have plenty of room there (you use about one fourth of the ADCs input range then, so you get 8 bits of effective resolution over your temperature range).
If you want to improve resolution, use a 3.3V low-noise regulator to create both Vref for the AVR, and Vcc for the AD8459 (it can run with this voltage). That way you can be sure the voltage from the thermocouple amplifier never exceeds the reference voltage.
But you also can use a zener diode to clamp the voltage of the amplifier. When looking at e.g. the ATMega16 data sheet (you did not specify which AVR you use) it has an input resistance am 100MOhm, and states that an input impedance of less than 100kOhm is suggested. So the clamping will not have any effect as long as R1 in the schematic above is small enough. And 10kOhm would be perfectly OK - the amplifier then needs to source 0.5mA.
Using an external ADC is another solution. If you can afford the board space and the additional components, it seems even like the best solution. Look for an ADC with a reference of 2V which also can stand inputs up to its Vcc, then you are fine.
I personally would go with the 3.3V LDO solution. You might need a stable and noise-free reference voltage anyway, so why not using it to solve other problems as well?
A pot doesn't break KCL or KVL, and as you say it's just a variable resistor. Usually it controls current by converting a voltage on the pot to a current. That's just Ohm's law- The current is the voltage across the resistance divided by the (variable) resistance.
There's no law that says a series circuit has a constant current, only that elements in series carry the same current.
Best Answer
My understanding is that the thermocouple voltage is strictly a function of the temperature. Apparently this is called the Seebeck Voltage. Like any voltage source there is an internal resistance associated with it due to "real world" effects. The short-circuit current will be determined by that internal resistance by the usual Ohm's Law calculation I_ss = V(temperature) / R_internal.