32 kHz crystal not operating as expected
There are two main things going on:

You don't have enough load capacitance.

You don't understand load capacitance.



Imagine one side of a crystal being driven with a sine wave at the crystal frequency.  This signal is low impedance.  The load capacitance is that capacitance you put on the other side of the crystal to cause a 180&deg; phase shift.

The phase shift of such crystals varies quickly as a function of the frequency at the crystals operating frequency.  Since the phase as a function of frequency is very steep right at the operating frequency, this is a good thing for the driving circuit to use to ensure the crystal is operating at the intended frequency.  These types of circuits oscillate optimally when the crystal shifts the input phase by 180&deg;.  Since only a little frequency change messes that up, the resulting oscillation is very close to the crystal's intended frequency.

Now back to your circuit.  The big clue is that things work when you put a scope probe on the oscillator input pin.  What that is doing is adding capacitance on the output side of the crystal.  Apparently, with the setup you have, the additional scope probe capacitance causes the crystal to shift the phase the appropriate amount for the system to oscillate.  If you add more capacitance yourself to the crystal output only, you replicate the effect of the scope probe and things will work.  Try another 10 pF or so for starters.

Don't use formulas you find on the other end of the internet without understanding them.  The equation you show makes a bunch of assumption, some of them not valid.  Unfortunately there is a lot of conventional stupidity out there regarding crystals.

The crystal by itself is only a two-terminal device and doesn't "know" anything about your circuit ground.  Ultimately, the load capacitance is what is across its terminals.  The conventional stupidity therefore says to use two equal capacitors on each side of the crystal to ground.  Since these are in series, each need to be twice the desired capacitance.  However, whatever stray capacitance to ground you think there is on each side of the crystal needs to be subtracted from these capacitances.

The problem with the conventional stupidity is that it ignores the impedance of the crystal driver output.  Consider the extreme case where that is 0.  In that case, the capacitance added on the input side of the crystal is completely irrelevant, since it is in parallel with the 0 impedance of the driver.  The load on the crystal is then only the capacitance on its output.

Do some math.  The impedance of 6 pF at 32.8 kHz is 810 k&Omega;.  Now the impedance of the crystal driver is certainly not zero, but quite likely significant relative to 810 k&Omega;.

Consider what each of the caps really do.  The one on the input loads the crystal driver.  The main purpose of that is to attenuate some of the harmonics coming out of the driver.  This beats on the crystal less, and makes it less likely that the whole system will oscillate at a harmonic.  Crytals have complex transfer characteristics.  They can have some of the same characteristics at harmonics as they do at the intended operating frequency.  Some crystals are cut so as to deliberately enable use at harmonics, called overtone mode in the industry.

The capacitance on the output is the true "load" capacitance.  Its reactance works against that of the crystal to phase shift the result the right amount at the right frequency.

In your case, the crystal is rated for 6 pF load, and that's what you put on its output.  That should have worked.  My guess what is happening is that the cap on the input of the crystal, really on the output of the crystal driver, also caused a phase shift that worked against that of the load cap.  Just as a test, try removing the cap on the crystal input and leave the 6 pF on its output.  It would be nice to see the waveshape on the crystal input then, but even a 10x scope probe might change it.  Try it anyway, but make sure the scope probe is set to the highest impedance, therefore the lowest capacitance, possible.