How to calculate the value of the thermistor based on the output voltage of this circuit
This circuit can be split into four sections


Reference generation
Wheatstone bridge
Differential amplifier
1st order filter
non-inverting amplifier


Reference Generation

A simple voltage divider from Vcc and buffered to provide a low-impedance source to the bridge:  

U2B V+ pin sits at 0.5Vcc and thus Vref ~ 2.5V  (accuracy of the rail and resistors dependant)

Wheatstone Bridge

The resistors around the Thermistor (R13,14,15) are 10k so it is safe to assume that the Thermistor is a 10k at some reference temperature. 

The thevenin equivalent at this operating point for nodes A &amp; B is therefore 5k with the voltages at these nodes being 1.25V 







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Differential Amplifier

Assuming you forgot the 20k from V+ to GND, the differential amplifier stage has a gain of \$ \frac{ 20k}{10k + 5k} \$ = 1.33







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1st order filter

This will provide some attenuation at non-DC temperature changes, but assuming the rate of change of thermistor resistors in &lt;&lt; the 3dB point of the 1st order filter, this will not change the gain of the circuit (except for a leakage component)

non-inverting amplifier

The final stage is a classic non-inverting amplifier with a gain of 11 ( \$\frac{100k}{10k} + 1\$ )

Overall gain is therefore 14.63 when the temperature is such that the thermistor presents 10k resistance.

As the temperature changes then the  bridge impedance will change. This will produce a non-constant impedance and thus the gain of the differential amplifier will also vary with temperature. If this is acceptable this is a cheap implementation. Creating an instrumentation amplifier would produce a more constant impedance and thus gain response. You have an unused OPAMP so this is a possibility