I've studied the LEM datasheet and had a quick look at the referenced answer so I may have missed some detail but I think you can simplify it.
Figure 1. Extract from the LEM datasheet showing the maximum \$ R_M \$ the device can drive under various conditions.
Let's go with 100 Ω for now. You can increase it if it suits. This will drop 1.1 V at 11 mA which is good for our application - but bear in mind that a current spike or surge could raise this beyond your ADC's maximum.
simulate this circuit – Schematic created using CircuitLab
Figure 2. Biasing the \$ R_M \$ voltage reading to mid-ADC range.
Fortunately for what we want to do next \$ R_M \$ is a low value and the ADC has (he said without checking) a high input impedance.
- If \$ V_M \$ (measured voltage on \$ R_M \$ = 0 then we want half supply voltage on the input. We can do this with a 1:1 potential divider. We'll pick 47k resistors as these are not so high as to introduce noise susceptibility but 500 times higer than \$ R_M \$ and so should introduce a measurement error of less than 0.2%.
- If \$ V_M \$ were to rise to 3.3 V the ADC input would also be 3.3 V. (This would be 33 mA output.)
- If \$ V_M \$ were to fall to -3.3V the ADC input would be held at the midpoint of +3.3 and -3.3 V = 0 V.
- At \$ V_M \$ = 0 the ADC input will be 1.65 V.
- At +11 mA the ADC input will go up 1/3 of the way from midpoint to +3V3. At -11 mA the input will go down 1/3 of the way from midpoint to 0 V.
You'll need to work out if this gives you adequate resolution and balance the trade-off between sensitivity and overload headroom.
It might be a good idea to add protection diodes to the ADC input - one from ground up to the input and another from the input to V+.