How do I momentarily ground a 3.3V signal (with a pull-up resistor) when power is first applied?
This is the kind of thing a reset supervisor often does for you.  It's job is usually to hold a reset line low (asserted) until the power has been stable for some particular time.  It can be used for a purpose like this too.

However, what you describe is simple and can can be done with a few discrete parts.  Here is one example:



When the power has been off for a while, C1 is discharged.  When power comes on, the instantaneous voltage on C1 doesn't change, so the top of R1 goes to POWER.  This turns on the transistor and holds OUT low.  As current flows thru R1, C1 charges up exponentially.  During this time, the transistor is on, so the base voltage is roughly 600 mV.  Eventually C1 charges up enough so that its bottom voltage after being divided by R1 and R2 is not
enough to keep the transistor on, and it turns off, thereby releasing OUT to high impedance.

The purpose of R2 is to eventually bring the base voltage to 0, keeping the transistor solidly off.  It also makes the time until the transistor turns off more predictable because it relies less on the unpredictable base current when it is barely enough to keep the transistor on.  This also speeds up the on to off transition.  The actual time is somewhat unpredictable, and depends somewhat on the POWER level.

D1 is there to actively discharge C1 whenever the power goes low.  Without D1, C1 will eventually discharge thru R1 and R2 when the power is off.  D1 forces the startup timeout to begin anew even after a short no-power glitch.  Depending on your situation, you may be able to remove D1.