RTC circuit supercapacitor charge/discharge doubts
What value of resistance should I consider in practice? 150 ohms (2.5 minutes)? Or 300 ohms (5 minutes)?

The ESR of the cap plus external resistance.  You should model the cap as a &quot;pure&quot; capacitance with a 150\$\Omega\$ resistor in series.

Do you know some online calculator where I could estimate that time of operating from VBAT pin after power-off of the board and capacitor being 95% charged?

Well this is on line, presumably you have a calculator, so bear with me a moment...
The Math
\$V_{cap} = \frac{q_c}{C}\$, where \$q\$ is the charge on the cap in coulombs, and \$C\$ is the capacitance.  And \$q_c(t) = \int i_c dt\$.  It's probably safe to assume that the RTC clock current consumption is constant or goes down as voltage goes down, but your should check this.  If true, then the change in charge is just equal to current times time: \$\Delta q_c = \Delta i_c t\$.  Then \$\Delta V_c = \frac{\Delta q_c}{C} = \frac{i_c \Delta t}{C}\$.
If you want to go from 3.1V down to 1.3V, then \$\Delta V_c = \mathrm{3.1V - 1.3V = 1.8V}\$.  Then doing a bit of algebra gets \$\Delta t = \frac{C \Delta V_c}{i_c}\$.  Using your numbers, \$\Delta t\$ is then \$\mathrm{\frac{1.8V\ 0.33F}{1.2\mu A} = 495 \cdot 10^3}\$ seconds.
However -- Shotkey diodes leak, particularly at high temperatures.  You may want to revisit your diode choice.