The best thing you can do is to simply add a DC offset to your signal. Instead of having it be 0V ±80mV, have it 1.65V ±80mv. All you need is a capacitor and two resistors:
simulate this circuit – Schematic created using CircuitLab
However, 80mV is not going to have much resolution unless you reduce your reference voltage considerably, and in that case you'd need to reduce the 3.3 down to the same (or around) the reference voltage you use.
But, just feeding your signal directly into the ADC will give you the positive 80mV signal which you can use a smaller reference voltage with directly with no problem - after all, the datasheet shows:
Voltage on Input Pins
with Respect to Ground (except VBUS)...............-0.3V to +4.0V
So you can provide down to -300mV on an analog input before it potentially becomes a problem.
This is because all the IO pins have ESD protection diodes, which look like this:
simulate this circuit
Those diodes have a forward voltage of about 300mV, and so anything below -300mV will cause the lower diode to start conducting. When that happens too much current can flow and the diode risks burning out. But you can add further protection by simply adding a small resistor (1kΩ is good) in series with the IO pin to limit the current in the case that the voltage goes below -300mV and protect the internal circuitry.