Effect of Pi Filter on the Output of LDO
If there is not other decoupling capacitance across your 3.3 V supply (and taking a beginner's view of it)...

Your microcontroller is a rapidly switching load. It draws a DC current plus spikes of current drawn when the logic transistors within it switch in response to its internal logic clock. You don't say but I imagine this is in the tens of megahertz.

The regulator cannot react to changes of output load at these speeds. Therefore local decoupling capacitors provide current to the load during these spikes of draw and charge up in between them. The voltage across the load is no longer tightly coupled to the regulator response. The capacitors have 'de-coupled' the load voltage from it somewhat.

The inductor will oppose this switching current flow. It will have a high impedance in response to the load current changing sharply when it spikes, so the voltage across the load will fall. This is why a Pi filter in a voltage supply uses an output capacitor. A multimeter measures average voltage so it might not show these dips in supply voltage very well. A 'scope will show it much better.

As a rule, keep series inductance out of power supplies wherever possible. Only put it in where needed, rather than just as an idea. (That's true of every component in any circuit.) If you need to filter high supply frequencies from your reaching your circuit, put the series inductor on the linear regulator supply. But ensure your regulator has suitable input and output decoupling capacitance for the loads it has.