In theory this could work (though you would need another diode to load the transformer on the negative half cycles) since the CT is approximately a constant current source when appropriately loaded, however it would have to be a different (much larger core) CT.
As you can see below, with the give CT there will be very little signal available at the output.
simulate this circuit – Schematic created using CircuitLab
The recommended maximum burden for this transformer to meet specs is 7.5\$\Omega\$, which means that it can only output 300mV RMS at 100A in. The diode drop is included in that maximum output voltage, which leaves you with very little signal even with a Schottky diode.
Your best bet probably is to load the part with 7.5\$\Omega\$ or so, giving you 300mV@100A, and amplify the AC signal to a couple volts (of the order of 10:1) with an op-amp then apply an AC-coupled precision rectifier (full wave or half wave). This can be done with a single supply if you AC-couple the amplifier or bias it to the middle of your 3.3V supply.
Below (from a TI document) is another way to do with a generated -230mV supply (so it can operate without an external negative supply). You can modify the feedback of the amplifier U2A (by increasing the value of R2- to 10K or 15K) to get the required output voltage, and of course you need the burden resistor R4 to be the correct value- less than 140mV RMS, so about 3.5 ohms for 100A full scale. A 5V supply is shown but it will operate from 3.3V.
Edit: To respond to a query in the comments- below is an LTSpice simulation to aid in understanding my first circuit above. Current is steered by the diodes either through the 1R resistor via D1 for positive current or through D2. The 55mV peak voltage can be low-pass filtered and amplified.
The simulation has a 1:2500 turns ratio ideal CT measuring 100A RMS from a 2.2 ohm resistor on a 220VAC RMS supply. The presence of the diodes has a negligible effect on the response.