Detrmining open loop gain of an amplifier in LT-spice using a transient simulation
You could break apart into stages, and consider separately. First, the sampling is 925Hz, so anything beyond f0/2 is simply ignored. Second, there are two paths, dictated by each of the two pairs of switches, \$Q\$ and \$\overline{Q}\$.

The first stage would be the input RC, that's easy: fc~528kHz, consider flat. When \$Q\$ is active, input is active, else output is active; both go directly into the JFETs.

Then there's the differential nJFET together with A1 (they share the feedback network). Since the JFETs don't have capacitors around them, and since their parasitics would, most likely, go well beyond Nyquist, they're considered to have flat response, while the feedback network says you have a pole at \$\frac{1}{2\pi 10\text{k}1\mu}\approx 16\text{Hz}\$, with \$Av=\frac{10\text{Meg}}{10\text{k}}=60\text{dB}\$ and a zero at \$\frac{1}{2\pi 10\text{Meg}1\mu}\approx 16\text{mHz}\$, \$Av=0\text{dB}\$. Or write its transfer function as a shelf highpass (with R2=10Meg, R1=10k, C=1\$\mu\$):

$$H_1(s)=\frac{R_2}{R_1}\frac{s}{s+\frac{1}{R_1 C}}$$

There is at least a pole of the opamp which is beyond Nyquist, ignored.

The last stage is the integrator plus or minus the 1\$\mu\$F cap + switches. Their RDSon is considered too low compared to RDSoff. If \$Q\$, A2's input is grounded, its output will be the discharge of the feedback cap. If \$\overline{Q}\$, input is a passive highpass with C and RDSon, terminated with RDSoff which is, normally, much higher than the 240k, thus the output will be influenced by the integrator's input, which will also change the overall transfer function. At this point, I  call LTspice to help and it turns out that the relevant transfer function is that of a highpass with fc well below 1mHz (~159\$\mu\$Hz, considering RDSoff=1G\$\Omega\$), together with a shelf lowpass with only the pole being below Nyquist, at ~325Hz (quite close to 462.5Hz), so you could write:

$$H_2(s)=\frac{s}{s+\frac{1}{R_{DSoff}C}}\frac{\frac{1}{R_1 C}}{s+\frac{1}{R_1 C}}$$

The overall open loop gain should be \$H_1(s)H_2(s)\$, depending on which path is active, \$Q\$, or \$\overline{Q}\$.