At 20 kHz, the phase shift of the SMA connector is negligible. The length of the signal path is maybe 5 mm. The dielectric constant of the PTFE (Teflon) material in the SMA connector is about 2.0. So the delay through the connector is about 23 ps. That's about 20 microradians of phase delay at 20 kHz.
Even at higher frequencies, the delay through the SMA connector is probably quite small compared to the delay through the cable that's connected to the connector, which you've said nothing about.
If you were working at a higher frequency where 20 ps made a difference, then you'd have a problem. Because the delay induced by a connector like this depends on the footprint on the PCB where you mount it as well as the construction of the connector itself. And because the two sides of the connector aren't the same type of waveguide.
The best way to determine the characteristics of the connector would probably be to build two additional "test coupons" onto your pcb. Each coupon would be a length of trace with a connector on each end. The two traces would be different lengths. By measuring the S-parameters of the two coupons with a network analyzer, you'd have enough information to cancel out the affect of the traces and determine the characteristics of the connector.
I think that writing the loop equations would be easier.
The loop equations for first two loops:
$$I_1(Z+R) = V_{in}$$
$$- I_1R +I_2(Z+2R) -I_3R = 0$$
Where \$Z=\dfrac{1}{Cs}\$. From this:
$$I_2(Z+2R) -I_3R = V_{in}\frac{R}{R+Z}\tag1$$
The remaining two loop equations:
$$- I_2R +I_3(Z+2R) -I_4R = 0\tag2$$
$$- I_3R +I_4(Z+2R) = 0\tag3$$
Expressing in matrix form:
$$\left[\begin{array}{ccc}
&Z+2R &-R &0 \\
&-R &Z+2R &-R \\
&0 &-R & Z+2R
\end{array}\right] \left[\begin{array}{c}
I_2\\
I_3\\
I_4
\end{array}\right] = \left[\begin{array}{c}
\frac{V_{in}R}{R+Z}\\
0\\
0
\end{array}\right]$$
Now by Cramer's rule:
$$ I_4 = \frac{\left|\begin{array}{ccc}
Z+2R &-R &\frac{V_{in}R}{R+Z} \\
-R &Z+2R &0 \\
0 &-R & 0
\end{array}\right|}{\left|\begin{array}{ccc}
Z+2R &-R &0 \\
-R &Z+2R &-R \\
0 &-R & Z+2R
\end{array}\right|}$$
$$V_{out} = I_4\times R$$
From this the transfer function can be calculated. Gain and phase shift can be calculated from transfer function. (substitute \$Z=\frac{1}{jwC}\$)
Best Answer
With a circuit like this one, there are two options: the brute-force approach or the FACTs. For the first one, you can apply the superposition theorem with the input voltage alternatively biasing \$R_1\$ and \$R_3\$ or use the more mathematical KVL-/KCL-based approached described by Jan in his answer.
The second approach that I favor most of the time consists of determining the time constant of this circuit as described by the fast analytical circuits techniques or FACTs. As described in my book on the subject, the principle is simple: determine the time constant involving \$C_1\$ when the excitation is reduced to 0 V (\$V_{in}\$ is replaced by a short circuit) and when the response is nulled. We can start by setting \$s\$ to 0 (dc gain) and see what gain does this circuit exhibit. The below drawing shows the step for this dc analysis. If you do the maths ok (via superposition for instance), you find a gain of 1 regardless of the resistors values:
You then carry on with turning the source off and determining the resistance "seen" from \$C_1\$ terminals. It's easy in this case as it is \$R_1\$ and you have \$\tau_1=R_1C_1\$. Because with a 1st-order circuit the pole is the inverse of the time constant. The pole position is immediate and equal to \$\omega_p=\frac{1}{R_1C_1}\$. For the zero, either you determine the value of \$R\$ when the output is a null or you determine the gain of the circuit at high frequency e.g. when \$C_1\$ is replaced by a short circuit: \$H^1=-\frac{R_4}{R_3}\$. Applying this result by following the guidelines you have a zero located at \$\omega_z=-\frac{R_3}{R_4R_1C_1}\$ and this is a right-half-plane zero (RHPZ).
Since the pole and the zero are located at the same position, they neutralize each other in magnitude and the overall gain is 1 (0 dB) along the frequency axis. However, because of the zero located in the right half-plane, its phase response lags rather than leading as with a classical LHP zero. As a result, it combines with the naturally-lagging response of the pole and brings the phase from 0° down to -180°:
Because \$R_3\$ and \$R_4\$ are of equal values, the pole and the zero solely depend on \$R_1\$ and \$C_1\$. Considering a negative zero, the phase of the expression is thus classically computed as:
This is an all-pass filter which creates a delay whose value depends on the value of \$R_1\$ and \$C_1\$. If you push the maths a little more, this is a way to create a 1st-order Padé approximant of \$e^{-sT}\$ which is a pure delay.