What is the exact value of $\eta(6i)$? Let $\eta(\tau)$ be the Dedekind eta function. In his Lost Notebook, Ramanujan played around with a related function and came up with some of the nice evaluations,
$$\begin{aligned}
\eta(i) &= \frac{1}{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(2i) &= \frac{1}{2^{11/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(3i) &= \frac{1}{2\cdot 3^{3/8}} \frac{1}{(2+\sqrt{3})^{1/12}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(4i) &= \frac{1}{2^{29/16}} \frac{1}{(1+\sqrt{2})^{1/4}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(5i) &= \frac{1}{2\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(6i) &=\; \color{red}{??}\\
\eta(7i) &= \frac{1}{2\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(8i) &= \frac{1}{2^{73/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(16i) &= \frac{1}{2^{177/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\end{aligned}$$
with the higher ones $>4$ added by this OP. (Note the powers of $2$.)

Questions: 
  
  
*
  
*Similar to the others, what is the exact value of $\eta(6i)$?
  
*Is it true that the function,
  $$F(\sqrt{-N}) = \frac{\pi^{3/4}}{\Gamma\big(\tfrac{1}{4}\big)}\,\eta(\sqrt{-N}) $$
  is an algebraic number only if $N$ is a square?

P.S. It seems strange there is a function that yields an algebraic number for square input $N$ and a transcendental number for non-square $N$. (Are there well-known functions like that?) For an example of non-square $N$, we have,
$$\eta(\sqrt{-3}) =  \frac{3^{1/8}}{2^{4/3}} \frac{\Gamma\big(\tfrac{1}{3}\big)^{3/2}}{\pi} = 0.63542\dots$$
and $F(\sqrt{-3})$ seems to be transcendental.
 A: The answer to the second question is very simple.
Assume the complete elliptic integral of the first kind 
$$
K(x)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-x^2\sin^2(\theta)}}=\frac{\pi}{2}\cdot {}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right).
$$
Then if we define the elliptic singular modulus $0<k_r<1$ to be the solution of 
$$
\frac{K\left(\sqrt{1-k_r^2}\right)}{K(k_r)}=\sqrt{r}\textrm{, }r>0,
$$
we know that $k_r$ is algebraic for $r$ positive rational. Assume the notation: 
$$
K[r]:=K(k_r)\textrm{, }r>0
$$
Also for $r_1,r,N$, such that $N$ positive integer, $r\in Q^{*}_{+}$, with $r_1=N^2r$, we have
$$
\frac{K[N^2r]}{K[r]}=M_N(r)
$$ 
The function $M_N(r)$ is called: "the multiplier" and for $r,N$ as above it is an algebraic number.
The Dedekind eta function is just
$$
\eta(z)=\eta_1(q):=q^{1/24}\prod^{\infty}_{n=1}\left(1-q^n\right)\textrm{, }q=e^{2\pi i z}\textrm{, }Im(z)>0.\tag 1
$$ 
We want to find $\eta(i\sqrt{r})$, $r>0$. For this we define the Ramanujan eta function
$$
f(-q):=\prod^{\infty}_{n=1}\left(1-q^n\right)\textrm{, }q=e^{-\pi\sqrt{r}}\textrm{, }r>0.\tag 2
$$
The function $f(-q)$ can evaluated at $q=e^{-\pi\sqrt{r}}$, $r>0$ as (see [W,W] Chapter 21 pg.488):
$$
f(-q)=\frac{2^{1/3}}{\sqrt{\pi}}q^{-1/24}(k_r)^{1/12}(k'_r)^{1/3}K[r]^{1/2}\tag  3
$$
Assume that $q_1=e^{-\pi\sqrt{r_1}}$, with $r_1=N^2r$ and $GCD(N,r)=1$, ($r_1,N,r$ are integers, $N>1$ and $r$ is square free), then from (1),(2),(3) we have:
$$
\eta(i\sqrt{r_1})=\eta(i\sqrt{N^2r})=q_1^{1/12}f(-q_1^2)=algebraic\cdot \frac{\sqrt{K[4r_1]}}{\sqrt{\pi}}=alg\cdot\frac{\sqrt{K[4N^2r]}}{\sqrt{\pi}}=
$$
$$
=alg\cdot \sqrt{M_2(r)}\sqrt{M_{N}(r)}\frac{\sqrt{K[r]}}{\sqrt{\pi}}
$$ 
Hence if $r_1$ is a square (which means $r=1)$, then
$$
\eta(i\sqrt{r_1})=alg\cdot\sqrt{\frac{K[1]}{\pi}}=alg\cdot \frac{\Gamma\left(\frac{1}{4}\right)}{\pi^{3/4}}
$$
Q.E.D.
Note:
$k'_r=\sqrt{1-k_r^2}$. $k'_r$ is called the coplementary modulus.
References
[W,W]: E.T. Whittaker and G.N. Watson. 'A course on Modern Analysis'. Cambridge U.P. 1927.
A: While giving this and this answer I got all the ingredients necessary to solve the current problem.

Consider the nome $q=e^{-\pi} $ so that elliptic modulus is $k=2^{-1/2}$ and the function $f(q) $ defined by $$f(q)=q^{1/6}\prod_{n=1}^{\infty}(1-q^{4n})\tag{1}$$ which is expressed in terms of modulus $k$ and complete elliptic integral $K=K(k) $ as $$f(q) =2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{2}$$ (see first answer linked above).
If $l$ is the elliptic modulus corresponding to nome $q^{3}$ then we have $$\eta(6i)=f(q^{3})=2^{-2/3}\sqrt{\frac{2L}{\pi}}(l^2\sqrt{l'})^{1/6}\tag{3}$$ The values $$\sqrt{\frac{2L}{\pi}} =\frac{(27+18\sqrt{3})^{1/4}\Gamma(1/4)} {3\sqrt{2}\pi^{3/4}} ,l=\frac{(\sqrt{3}-1)(\sqrt{2}-\sqrt[4]{3})}{2},\sqrt{l'}=\frac{\sqrt{2}+\sqrt[4]{3}(\sqrt{3}-1)}{2^{5/4}}$$ are available from the second answer linked above which help us to get the value of $\eta(6i)$ in explicit form. The calculations are somewhat lengthy but can be performed using pen and paper. 
A: After persevering with a Mathematica session, I found that $F(6i)$ is the root of $96$-deg eqn (no wonder it was hard to find!) but could be prettified as,
$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \left(\frac{5-\sqrt{3}}{2}-\frac{3^{3/4}}{\sqrt{2}}\right)^{1/6}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$
However, the second question is still open.

$\color{blue}{Update}$: Four years after this question was asked, Giuseppe Manco found a more elegant factorable formulation in this long post,
$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \big(2-\sqrt3\big)^{1/24}\, \big(\sqrt2-\sqrt[4]3\big)^{1/4}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$
A: Since we know the value of $\eta(3i)$, the point is just to compute the value of the product:
$$ \prod_{n\geq 0}(1+e^{-6\pi n})=\exp\sum_{n\geq 0}\log\left(1+e^{-6\pi n}\right)=\exp\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds$$
where:
$$\sum_{n\geq 0}\int_{n}^{n+1}\frac{6\pi n}{1+e^{6\pi s}}\,ds = \int_{0}^{+\infty}\frac{6\pi s\,}{1+e^{6\pi s}}-\int_{0}^{+\infty}\frac{6\pi\{s\}}{1+e^{6\pi s}}$$
and the first integral in the RHS equals $\frac{\pi}{72}$ by the residue theorem, while expanding the fractional part as its Fourier series, $\{s\}=\frac{1}{2}-\sum_{n\geq 1}\frac{\sin(2\pi n s)}{\pi n}$, we get:
$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{6\pi\{s\}\,ds}{1+e^{6\pi s}}&=&\frac{\log 2}{2}-\sum_{n\geq 1}\int_{0}^{+\infty}\frac{6 \sin(2\pi n s)}{n(e^{6\pi s}+1)}\,ds\\&=&\frac{\log 2}{2}-\sum_{n\geq 1}\frac{6}{n}\sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\sin(2\pi n s)\,e^{-6\pi m s}\,ds\\&=&\frac{\log 2}{2}-\frac{3}{\pi}\sum_{n\geq 1}\sum_{m\geq 0}\frac{(-1)^m}{9 m^2+n^2}\\&=&\frac{\log 2}{2}-\frac{\pi}{2}-\frac{3}{\pi}\sum_{m,n\geq 1}\frac{(-1)^m}{9m^2+n^2}\end{eqnarray*}$$
and the last series just depends on the number of ways to represent a positive integer $\not\equiv 2\pmod{3}$ through the binary quadratic form $n^2+9m^2$: it is, with minor manipulations, just a Dirichlet convolution. I have just applied the same techniques of this answer, just in reverse. 
This shows a clear connection between the evaluation of the Dedekind eta function at quadratic irrationals and the class number problem: $\eta(\sqrt{-N})$ depends on $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$, where $r(n)$ counts the number of ways to represent $n$ as $a^2+Nb^2$. If $N$ is a square or $a^2+Nb^2$ is the only reduced quadratic form of discriminant $-4N$ (class number one) we may explicitly compute such series, and it turns out that $F(\sqrt{-N})$ is an algebraic number. Otherwise, $\sum_{n\geq 1}(-1)^n\frac{r(n)}{n}$ is not even a convolution of Dirichlet series, hence your conjecture is very likely to hold.
Ultimately, the computation of $\eta(6i)$ can be carried on by recalling that:
$$j(\tau)=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8+2^8\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^{3} $$
and by computing the Klein $j$-invariant $j(3i)$. The Wikipedia page gives:
$$ j(3i) = \frac{1}{27}(2+\sqrt{3})^2(21+20\sqrt{3})^3.$$
A: This is a question about the findings of this post as well as this one. If we like the question I can ask it as a new question for the site but it may be that this type of thing is trivially shown not to be the case with some counterexample or maybe it follows quickly: I am not sure. 
Let $\alpha = (-\frac{3}{4})!\pi^{-\frac{3}{4}}$ then we have seen that for the set $S=\{\frac{1}{n} : n\in \mathbb{N} \}\cup \mathbb{N}$ we have that 
$\eta (Si)\subset A[\alpha]$. So a natural question might be: is it the case that $\eta(\mathbb{Q}i)\subset  A[\alpha]$?
I am already changing my completely unverified conjecture: 
Maybe it's more like there is some transcendental number $t_m$ we can associate with each $m\in \mathbb{N}$ such that $\eta(i\{x:x \in \{ \frac{m}{n}\} \cup \{ \frac{n}{m} \} \})\subset A[t_m]$? What we know is that this $\eta$ function needs to respect multiplicative inverses in someway. 
