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Applying overvoltage to analogue inputs is the most likely way that I can see for damaging our Arduinos, particularly for the current sensor.  Although most of us only need to measure currents up to around 13A, household consumption can far exceed this level.  At such times, the CT's output signal is likely to be well in excess of 5.5V, so we're totally reliant on protection diodes, both internal and external.

so we're totally reliant on protection diodes, both internal and external  aided, in the case of the emonTx, by the impedance of the bias chain which should just about limit the current to a safe value.

But a series resistor right next to the input pin, after any external clamping diodes, would pretty much ensure safe operation under normal conditions. The full analysis needs to take the source impedance of the damaging voltage into account, together with all the surrounding circuitry.

Why does the series resistor need to be after the protection diodes rather than before them?  My gut feeling is that this arrangement leaves the protection diodes more vulnerable to being damaged. 

Is it likely that the Arduino's input circuitry would be damaged while the protection diodes are still intact?

The underlying assumption is that you have rated the external protection diodes to withstand the possible fault current. Because those diodes should limit the input to (Vs + 0.65) V or - 0.65 V (and assuming there's enough load so that the supply rail voltage does not lift in the case of a positive over-voltage), then the resistor only has to drop a few hundred (or even less) millivolts and hold the current down to a few milliamps to assure the internal protection diodes are not damaged. If there's a risk of a substantially higher voltage at the input (e.g. mains !), or enough current to lift the supply rail, then a second resistor before the external protection diodes, which might need to become fast zener diodes to common, would be necessary to limit the current. (But for a c.t. output with limited current capability, that's unlikely to be necessary).

Without the resistor after the external diodes, as the voltage dropped across the internal and external diodes is likely to be similar, there's no knowing how the current will share. Hence the resistor to ensure the external diodes, which can be suitably rated, take most of the current.

It is impossible to lay down rules of thumb to suit every situation. The input circuit has to be properly designed.