E=I*t=C*V
Consider a simple case, where there's only VDD and VDDA, VDD is the primary source, and VDDA is derived from it using a LDO which prevents reverse current flow, i.e. normally VDDA is lower than VDD by the LDO drop.
After switching off the primary power source, VDD starts to decrease, but if the load on VDDA is very small, the capacitors on VDDA may prevent VDDA from dropping quickly enough, and if VDD gets below VDDA, the internal structures in the chip start to behave like a diode and conduct current from VDDA to VDD.
The 1mJ requirement states, that in this case, the total energy which goes through those internal diode-like structures must stay below 1mJ. One way to interpret that is, that the current flowing in that way times the time it takes must stay below 1m A*s (i.e. 1mA*1s, or 10mA*100ms, etc.) Of course the current will change, so this is more an integral than just a plain multiplication.
Other way to look at it is, that if it's only the capacitors on VDDA which provide the reverse flow, then there's no more energy to be dissipated during that flow in the mcu than what's stored in those capacitors at the beginning of discharge. For example 3V on a 100nF capacitor represents 3V*100nF=300nJ.
JW
