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.

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,
$$\eta(6i) = \frac{1}{2\cdot 6^{3/8}} \big(2-\sqrt3\big)^{1/24}\, \big(\sqrt2-\sqrt[4]3\big)^{1/4}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$$
