For $|q|<1$ we consider the null Jacobi theta functions
$$
\theta_2(q):=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, }
\theta_3(q):=\sum^{\infty}_{n=-\infty}q^{n^2}\textrm{, }
\theta_4(q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}.
$$
For $q=e^{-\pi \sqrt{r}}$, $r>0$ the elliptic singular modulus $k=k_r$ is given by
$$
k_r=\left(\frac{\theta_2(q)}{\theta_3(q)}\right)^2.
$$
The complete elliptic integrals of the first and second kind are:
$$
K(x)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right)
$$
and
$$
E(x)=\frac{\pi}{2}{}_2F_1\left(-\frac{1}{2},\frac{1}{2};1;x^2\right).
$$
Then
Theorem.
Let
$$
f(-q):=\prod^{\infty}_{n=1}(1-q^n)\textrm{, }|q|<1,\tag 1
$$
then if $q=e^{-\pi\sqrt{r}}$, $r>0$, we have
$$
P(q^2)=1-24\sum^{\infty}_{n=1}\frac{nq^{2n}}{1-q^{2n}}=\frac{3}{\pi\sqrt{r}}+\left(1+k^2_r-\frac{3\alpha(r)}{\sqrt{r}}\right)\frac{4}{\pi^2}K^2.\tag 2
$$
Proof.
Let $q=e^{-\pi\sqrt{r}}$, $r>0$. Differentiating with respect to $r$ the relation
$$
\log\left(f(-q^2)\right)=\sum^{\infty}_{n=1}\log\left(1-q^{2n}\right)\textrm{, }|q|<1,
$$
and using (see [W,W] Chapter 21, Miscellaneous examples 10, pg. 488):
$$
f(-q^2)^6=\prod^{\infty}_{n=1}\left(1-q^{2n}\right)^6=\frac{2kk'K(k)^3}{\pi^3q^{1/2}},
$$
we get
$$
\frac{1}{6}\frac{d}{dr}\log\left(\frac{2k_rk'_rK^3}{\pi^3q^{1/2}}\right)=-2\sum^{\infty}_{n=1}\frac{nq^{2n-1}}{1-q^{2n}}\frac{dq}{dr}.
$$
After some calculations we arrive to (reminder: $P(q^2)=1-24P^{*}(q^2)$):
$$
\frac{1}{24}\left(\frac{\pi}{\sqrt{r}}+12\frac{1}{K}\frac{dK}{dk_r}\frac{dk_r}{dr}+\frac{4}{k_r}\frac{dk_r}{dr}+\frac{4}{k'_r}\frac{dk'_r}{dr}\right)=2q^{-1}P^{*}(q^2)\frac{q\pi}{2\sqrt{r}}
$$
Using the known relations (see [Ber3] Chapter 17 Entry 9 pg. 120 and [Ber2] Chapter 11 Entry 30 pg. 87-88):
$$
\frac{dk_r}{dr}=\frac{-k_r(k'_r)^2K^2}{\pi\sqrt{r}},
$$
$$
\frac{dk_r'}{dr}=\frac{k_r^2k_r'K^2}{\pi\sqrt{r}}
$$
and ([Borw,Borw] Chapter 1, Section 1.3, pg. 7-11):
$$
\frac{dK}{dk_r}=\frac{E}{k_r(k_r')^2}-\frac{K}{k_r},
$$
$$
\alpha(r)=\frac{\pi}{4K^2}-\sqrt{r}\left(\frac{E}{K}-1\right),
$$
we arrive to
$$
P^{*}(q^2)=-\frac{1}{24}+\frac{K^2}{6\pi^2}+\frac{K^2k_r^2}{6\pi^2}-\frac{\alpha(r)K^2}{2\pi^2\sqrt{r}}+\frac{1}{8\pi\sqrt{r}}.
$$
From this along with $P(q^2)=1-24P^{*}(q^2)$, we get the result. qed
Remark that $\alpha(r)$, $r>0$ is the elliptic alpha function in [Borw,Borw] book. It is proven in [Borw,Borw] that $\alpha(r)$ is algebraic number whenever $r$ is positive rational.
Proposition.
Let $r>0$ and $q=e^{-\pi\sqrt{r}}$, $K=K(k_r)$, then
$$
1-24\sum^{\infty}_{n=1}\frac{nq^n}{1-q^n}
=\frac{6}{\pi\sqrt{r}}+\left(1+k^2_r-\frac{6 \alpha(r)}{\sqrt{r}}\right)\frac{4K^2}{\pi^2}=
$$
$$
=\frac{6}{\pi\sqrt{r}}+s_1(r)\theta_3(q)^4,\tag 3
$$
where
$$
s_1(r):=1-\frac{6\alpha(r)}{\sqrt{r}}+k_r^2.\tag 4
$$
Proof.
From (1) setting $r\rightarrow r/4$ and using
$$
k_{r/4}=\frac{2\sqrt{k_r}}{1+k_r}\textrm{, }M_2(r)=\frac{1+k'_r}{2}
$$
and (see [Boew,Borw])
$$
\alpha(4r)=(1+k_{4r})^2\alpha(r)-2\sqrt{r}k_{4r},\tag a
$$
we get the result. qed
The relations you are looking for are (3) and (4).
Notes.
The upper half plane formulation can constructed if in place of $k_r$ we define for $q=e^{2\pi i \tau}$, $Im(\tau)>0$, the singular modulus:
$$
m(q):=\left(\frac{\theta_2(q)}{\theta_3(q)}\right)^2\textrm{, }q=e^{2\pi i \tau}\textrm{, }Im(\tau)>0\textrm{, }-1/2\leq Re(\tau)<1/2
$$
and $m'(q)=\sqrt{1-m(q)^2}$, then
$$
\theta_2(q)^2=\frac{2m(q)K(m(q))}{\pi}\textrm{, }\theta_3(q)^2=\frac{2K(m(q))}{\pi}\textrm{, }\theta_4(q)^2=\frac{2m'(q)K(m(q))}{\pi}.
$$
Then if also we have
$$
i\frac{K\left(m'(q)\right)}{K(m(q))}=2z
$$
see [Arm,Eb],[Borw,Borw]$\ldots etc$. You have to find the complex analog of elliptic alpha $\alpha(r)$.
References
[Arm,Eb]: J.V. Armitage, W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)
[Bag]: N.D. Bagis. 'On certain theta functions and modular forms in Ramanujan theories'. arXiv:1511.03716v2 [math.GM] 6 Dec 2017.
[Ber2]: Bruce. C. Berndt. 'Ramanujan`s Notebooks Part II'. Springer-Verlag, New York. 1989.
[Ber3]: B.C. Berndt. 'Ramanujan`s Notebooks Part III'. Springer Verlag, New York (1991).
[Borw,Borw]: J.M. Borwein and P.B. Borwein. 'Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity', Wiley, New York, 1987.
[W,W]: E.T. Whittaker and G.N. Watson. 'A course on Modern Analysis'. Cambridge U.P. 1927.
Continuing...
From relations (3) and (4) we get $q=e^{-\pi\sqrt{r}}$, $r>0$
$$
\left(E_2(q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\theta_3(q)^4}=1-\frac{6\alpha(r)}{\sqrt{r}}+k_r^2\tag 5
$$
If $r$ is positive rational, then $\alpha(r)$ and $k_r$ are algebraic numbers.
Moreover for any integer $N$ we can get easily the evaluation (in each case separate) of $k_r$ usng
$$
\left(\frac{256}{r_0^{16}}-r_0^8\right)^3=j\left(\frac{1+\sqrt{-N}}{2}\right),
$$
where $r=N-$integer and
$$
j\left(z\right)=\left(\left(\frac{\eta(z/2)}{\eta(z)}\right)^{16}+16\left(\frac{\eta(z)}{\eta(z/2)}\right)^8\right)^3
$$
is the known $j-$invariant. Then
$$
k_N^2=\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{16}{r_0^{24}}}
$$
An interesting article is [Broad] which reduces these calculations very considerably.
Another way is to evaluate it using
$$
j_r=j(i\sqrt{r})=\frac{256(k_r^2+(k'_r)^4)^3}{(k_rk'_r)^4}\textrm{, }k'_r=\sqrt{1-k_r^2}.\tag b
$$
The definition and evaluations of $\alpha(r)$ function are given in [Borw,Borw] and you must read the related theory (book).
The elliptic alpha function $\alpha(r)$ also has been introduced for evaluation of billions of digits of $\pi$. The Borwein bothers give in their book the underlined theory of $\pi-$formulas and algorithms.
Note also that [W,W] ($q=e^{-\pi\sqrt{r}}$, $r>0$):
$$
\prod^{\infty}_{n=1}(1-q^n)=2^{1/3}\pi^{-1/2}q^{-1/24}(k_r)^{1/12}(k'_r)^{1/3}K(k_r)^{1/2}\tag 6
$$
and
$$
\theta_3(q)=\sqrt{\frac{2K(k_r)}{\pi}}\tag 7
$$
and $k_r$ is algebraic number when $r-$positive raional.
Note also that the above two formulas have analytic continuations in upper half plane $Im(z)>0$.
References
[Broad]: David Broadhurst. 'Solutions by radicals at singular values $k_N$ from new class invariants for $N\equiv 3\textrm{mod}8$'. arXiv:0807.2976v3 [math-ph] 31 Jul 2008.
...Continuing
$$
A_r=\sqrt{-r}\left(E_2(q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\eta_1(q)^4}=\sqrt{-r}\frac{1+k_r^2-6\frac{\alpha(r)}{\sqrt{r}}}{2^{-2/3}(k_r)^{1/3}(k'_r)^{4/3}}\textrm{, }q=e^{-\pi\sqrt{r}}\textrm{, }r>0,\tag 8
$$
where
$$
\eta_1(q):=q^{1/24}\prod^{\infty}_{n=1}(1-q^n)\textrm{, }|q|<1.
$$
Your case is $q=e^{2\pi i z_N}$, where $z_N=\frac{N+i\sqrt{N}}{2}$, $N$ positive integer. Hence if $N$ is even, then $q=e^{2\pi i \left(N+i\sqrt{N}\right)/2}=e^{-\pi\sqrt{N}}$, $N$ positive even integer. Hence I can evaluate all your $N-$even cases.
1) If $N=r=4$, then
$$
k_4=3-2\sqrt{2},
$$
$$
k'_4=2\sqrt{-4+3\sqrt{2}},
$$
$$
\alpha(4)=6-4\sqrt{2}.
$$
Hence
$$
A_4=\ldots=0
$$
2) If $N=8=r$, then
$$
k_8=-2 \sqrt{2 \left(7+5 \sqrt{2}\right)}+4 \sqrt{2}+5,
$$
$$
k'_8=2 \sqrt{\sqrt{2 \left(799+565 \sqrt{2}\right)}-4 \left(7+5 \sqrt{2}\right)}
$$
$$
\alpha(8)=-8 \sqrt{41+29 \sqrt{2}}+26 \sqrt{2}+36
$$
Hence
$$
A_8=4i
$$
3) For $N=r=12$
$$
k_{12}=15-10 \sqrt{2}+8 \sqrt{3}-6 \sqrt{6}
$$
$$
k'_{12}=2 \sqrt{3 \sqrt{2 \left(4801-1960 \sqrt{6}\right)}+85 \sqrt{6}-208}
$$
$$
\alpha(12)=2 \left(132-94 \sqrt{2}+77 \sqrt{3}-54 \sqrt{6}\right)
$$
Hence
$$
A_{12}=6i4^{1/3}
$$
For $N=r=16$, we have
$$
k_{16}=-4 \sqrt{140+99 \sqrt{2}}+24 \sqrt{2}+33,
$$
$$
k'_{16}=2 \sqrt{6 \sqrt{69708+49291 \sqrt{2}}-8 \left(140+99 \sqrt{2}\right)}
$$
$$
\alpha(16)=4 \left(-4 \sqrt{15900+11243 \sqrt{2}}+252 \sqrt{2}+357\right)
$$
Hence
$$
A_{16}=12i\sqrt{2}
$$
For $N=r=28$, we have
$$
k_{28}=255-180 \sqrt{2}+96 \sqrt{7}-68 \sqrt{14}
$$
$$
k'_{28}=2 \sqrt{30 \sqrt{9321998-6591648 \sqrt{2}}+45798 \sqrt{2}-64768}
$$
$$
\alpha(28)=82464-58312 \sqrt{2}+31170 \sqrt{7}-22040 \sqrt{14}
$$
Hence
$$
A_{28}=54 i
$$
Where for all evaluations I have used [Borw,Borw] pg.172-173 and relation (a).
For odd $N$ we have $q=-e^{-\pi\sqrt{N}}$, which is very hard.
...CONTINUED
If $k_r=m\left(e^{-\pi\sqrt{r}}\right)$, with $m(q)$ is as below:
$$
m(q):=\left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^2,
$$
changing $q\rightarrow-q$, we have
$$
m(-q)=\left(\frac{i^{1/2}q^{1/4}\sum_{\scriptsize n\in\textbf{Z}}q^{n^2+n}}{\sum_{\scriptsize n\in\textbf{Z}}(-1)^nq^{n^2}}\right)^2=i\left(\frac{\vartheta_2(q)}{\vartheta_4(q)}\right)^2=i\frac{m(q)}{m'(q)}.
$$
Hence
$$
m(-q)=i\frac{\cdot m(q)}{m'(q)}\textrm{, }m'(q)=\sqrt{1-m(q)^2}.
$$
and from
$$
\frac{1}{\sqrt{1-x}}K\left(\frac{x}{x-1}\right)=K(x)
$$
we get
$$
K^{*}=K(m(-q))=m'(q)K(m(q))=m'(q)K.
$$
Hence
$$
\eta_1(-q)=\frac{e^{i\pi/24}2^{1/3}}{\sqrt{\pi}}\left(m(q)m'(q)\right)^{1/12}K(m(q))^{1/2}
$$
Hence if $N=r=$odd, we have $q=-e^{-\pi\sqrt{r}}$ and
$$
A_N:=\sqrt{-N}\left(E_2(z_N)-\frac{3}{\pi Im(z_N)}\right)\frac{1}{\eta(z_N)^4}=
\sqrt{-r}\left(P(-q)-\frac{6}{\pi\sqrt{r}}\right)\frac{1}{\eta_1(-q)^4}.
$$
But as Paramanand Singh noted, we have
$$
P(-q)=\left(\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+4m(q)^2-5\right).
$$
Also
$$
E=E(m(q))=\frac{\pi}{4\sqrt{r}K}+\left(1-\frac{\alpha(r)}{\sqrt{r}}\right)K.
$$
Hence for $N=r=$odd
$$
A_N=A_r=\ldots=\frac{e^{i\pi/3}4^{1/3}\left(-6\alpha(r)+\sqrt{r}(1+4m(q)^2)\right)}{\left(m(q)m'(q)\right)^{1/3}}\textrm{, }q=e^{-\pi\sqrt{r}}\tag 9
$$
Hence if $k_r$, $\alpha(r)$, for $r=3,7,11,19,27,43,67,163$ are known, then $A_r=A_N$ are known and hence answer to the evaluations of the case $z_N=\frac{N+\sqrt{-N}}{2}$, with $N-$odd. For all odd cases, since $h(-N)=1$ you can evaluate the $j-$invariant $j(i\sqrt{N})$ which (see [Broad]) have minimal polynomials of degree at most 3. Then using (b) you can find $k_r$.
For the elliptic alpha function, you have to search. With a quick view I find $r=3,7,27$ only (see [Borw,Borw] pg.173)
