The phase response is just the argument of the transfer function (just as the magnitude response is the absolute value).
The argument of a quotient is the argument of the numerator minus the argument of the denominator, i.e.,
$$
\phi = \operatorname{arg} \frac{j\omega}{j\omega+\frac{1}{RC}} = \operatorname{arg}(j\omega) - \operatorname{arg}(j\omega+\frac{1}{RC}) = \frac{\pi}{2} - \operatorname{atan2}(\omega, 1/RC )
$$
which is essentially just what you wrote.
Note that I did not write \$\operatorname{tan}^{-1}(y/x)\$ but \$\operatorname{atan2(y,x)}\$ because the former is only correct if \$x\$ is positive. It is incorrect when \$x\$ is zero or negative. In your case here, when \$\omega\$ is always positive that makes no difference, but I think it is good practice to use atan2.
1st stage is a 1.9995 kHz bandpass filter using a standard multiple-feedback topology. It has a Q of about 20 and a gain of about 40 dB.
2nd stage is a 2.009 kHz BP filter with a Q of about 25 and a gain of about 40 dB.
Total gain of first stage is 40 dB which makes a 10nV 2 kHz signal into 1 microvolt.
Bandwidth of 1st stage is \$\dfrac{f_C}{Q}\$ = 100Hz and for the purposes of noise calculation you can assume the bandwidth is 1.6x greater at 160 Hz.
\$E_{NOISE}\$ from ADA4004 is 1.8nV /\$\sqrt{Hz}\$. This means noise in a 160 Hz bandwidth is 22.8 nV
This is bigger than your signal (10nV) therefore this isn't going to be a great design.
Even if you took into account the Q of the 2nd stage, the bandwidth would only be halved i.e. a half power point would become a quarter-power point. Noise voltage would be about 16 nV and still a significantly bigger value than your signal. This is made worse by the 2nd stage being 9.5 Hz different to the 1st stage.
Best Answer
There are several mehods for calculating the closed-loop gain function of an opamp with feedback. I think, for the shown circuit the most simple way is as follows:
From H. Blacks famous formula - assuming ideal opamps with an open-loop gain of infinity - we know that the closed-loop gain can be written as
Acl=-Hf/Hr.
with (applying the superposition rule):
Hf (forward function)=Vn/Vi if Vout=0, and
Hr(return function, feedback factor)=Vn/Vout if Vi=0.
(Vn:voltage at the negative input terminal),
Both functions Hf and Hr can be found by applying the well-known voltage divider rule.