Electronic – the purpose of this NMOS transistor on an I/O pin

What do you want is called BMS(Battery Management System).

Basically you want to protect the battery so it can withstand among many other things: over/under current, over/under voltage. Since nowadays people are putting batteries almost everywhere, IC manufacturers responded appropriately. You will find tens of IC solutions for just the thing you need. Most popular being:

• TI - BQ76PL536(Tesla's choice)
• Linear Technology - LTC6804(BMW's choice)

Read the datasheet carefully and you should be able to create basic circuits. Go further, connect it with MCU capable of driving FET's as well as communicate with the IC and you will have very basic BMS structure.

However, if your budget could not afford one of the above IC, take analog multiplexer along with differential amplifier and use MCU's internal ADC for measurement. Luckily, you will have 12 bit resolution ~ 1-2mV measurement error(Li-Ion) after compensating

However, physically, there is no path between the input (gate) and the ground

Firstly, you do not need a capacitor connected to ground to create a pole in a system. Just the presence of the capacitor wherever it is connected is sufficient. A simple example is this high pass filter

This circuit has a transfer function given by:

$$H(s) = \frac{sCR}{1+sCR}$$ So we have a pole even though the capacitor is not connected to ground.

So, how are poles created?
One way to think about it is that the impedance of a capacitor (or an inductor) is given by \$1/sC\$ (\$sL\$), so whenever we have a these elements we will get a pole. Because it will result in the transfer function where the coefficients of terms with \$s\$ will be non zero.
Thus poles are created when we have energy storing elements present in a circuit. A capacitor will store energy in the form of voltage and the inductor in the form of current. The number of poles will be given by the number of indendent energy storing elements. For instance a capacitor in parallel with R1 in the above circuit will not create additional pole but if we cascade this system with another R and C, we will get a second order system.

Now, it should be clear that the miller capacitor will add to a pole. It's also easy to see why the capacitor is multiplied by gain. Since the capacitor is connected across an inverting gain amplifier, if the input of the amplifier rises by \$\Delta V\$ then the output would go down by \$A\Delta V\$. Thus the charge stored is \$\Delta Q = C(1+A)\Delta V\$. Thus, \$C_{eq} = \frac{\Delta Q}{\Delta V} = (1+A)C\$