The Dedekind eta function is denoted by $\eta(\tau)$, and is defined on the upper half-plane ($\Im \tau >0$). Put $\tau = i x$ where $x$ is
a positive real number. The function has the following representations:
$$\eta(ix)= e^{-\pi x/12} \prod_{n=1}^{\infty} (1-e^{-2\pi x n})
\\=\frac{2}{\sqrt{3}}\sum_{n=0}^{\infty} \cos\left(\frac{\pi}{6}(2n+1)\right)e^{-\pi x/12 \, (2n+1)^2}=\sum_{n \in \mathbb{Z}} (-1)^n e^{-\pi x/12 \,(6n+1)^2}.\tag{1}$$
It is not difficult to observe that when $x>0$, $\eta(i x)$ is a real number, and goes to $0$ when $x$ goes to infinity.
It also satisfies the functional equation $$\eta\left(\frac{i}{x}\right)=\sqrt{x}\,\eta(i x)\tag{2}$$
Moreover, the Jacobi triple product identity implies that
$$\eta^3(ix)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x(n+\frac12)^2}\\=\frac12\vartheta_1'(e^{-\pi x})=\frac12 \vartheta_2(e^{-\pi x})\vartheta_3(e^{-\pi x})\vartheta_4(e^{-\pi x})\tag{3}$$
Where $\vartheta_k$ are the Jacobi theta functions.
Now define
$$\large I(k)=\int_0^{\infty} \eta^k(ix)dx.\tag{4}$$
In his paper Some Integrals of the Dedekind Eta-function (2008) (arxiv link), Glasser shows
that $\displaystyle I(1)=\frac{2 \pi}{\sqrt{3}},I(3)=1$ (these follow directly from the above series represantations), and also gives the Laplace transform $$\int_0^{\infty} e^{-x y} \eta^3(i x)dx=\operatorname{sech}\sqrt{\pi y}\tag{5}$$
from which I can deduce (by setting $y=\pi(n+1/2)^2$, multiplying by $(-1)^n(2n+1)$ and summing) that
$$I(6)=\int_0^{\infty} \eta^6(ix)dx=\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{\cosh\left(\pi (n+\frac12)\right)}
\\=\frac12 \vartheta_2^2(e^{-\pi})\vartheta_4^2(e^{-\pi})=\frac{\pi}{4 \Gamma\left(\frac34\right)^4}.\tag{6}$$
By the way, note that the closed form for $I(1)$ is this question of @VladimirReshetnikov in disguise.
A numerical computation suggests that we also have
$$I(4)\stackrel?=\frac{2^{2/3} \pi}{3 \Gamma\left(\frac23\right)^3}.\tag{7}$$
$$I(8)\stackrel?=\frac23 \left(\frac{2^{2/3} \pi}{3 \Gamma\left(\frac23\right)^3}\right)^3.\tag{8}$$
Using the same procedure I did to evaluate $I(6)$ on another Laplace transform given by Glasser,
$$\int_0^{\infty} e^{-x y} \eta(i x)dx=\sqrt{\frac{\pi}{y}} \frac{\sinh 2\sqrt{\pi y/3}}{\cosh\sqrt{3 \pi y}},\tag{9}$$
I obtain $$I(4)=\int_0^{\infty} \eta^4(ix)dx=2 \sum_{n=0}^{\infty} (-1)^n \frac{\sinh\frac{\pi}{\sqrt{3}}(2n+1)}{\cosh\frac{\sqrt{3}\pi}{2}(2n+1)}\tag{10}$$
but I could not evaluate this sum in terms of elliptic functions as I did with $I(6)$. The formula for $I(8)$ is even more intriguing, as I cannot even see a reasonable route to start proving it.
Q1: Can we find a closed form expression for $I(n)$, at least for small integer $n$?
Q2: What is the closed form of $I(12)$?
Note that we have $$I(12)=\int_0^{\infty} \eta^{12}(ix)dx=\frac1{16} \int_0^{\infty}\vartheta_2^4(e^{-\pi x})\vartheta_3^4(e^{-\pi x})\vartheta_4^4(e^{-\pi x})dx=0.07552061383997469\dots$$
Based on the other results so far, I believe the closed form may involve the Gamma function.Bonus Q: What would be a way to prove the conjectural closed form for $I(4)$ or $I(8)$?
This is interesting to me because as can be seen in this post, a closed form for $I(12)$ may be used to find the closed form
of the integral $\int_0^{\infty} \vartheta_4^{12}(e^{-\pi x})/(1+x^2) dx$.
Another interesting aplication of these integrals is giving closed forms for beautiful lattice sums.
For instance, by expanding $\eta$ into its series representation, equation $(7)$ can be rewritten
$$\sum_{(a,b,c,d)\in \mathbb{Z}^4} \dfrac{(-1)^{a+b+c+d}}{(6a+1)^2+(6b+1)^2+(6c+1)^2+(6d+1)^2} = \frac{\pi^2}{18 \sqrt[3]{2}\,\Gamma\left(\frac23\right)^3},\tag{11}$$
which I deem visually pleasing. Similarly, equation $(6)$ can be rewritten
$$\sum_{(a,b,c)\in \mathbb{Z}^3} (-1)^{a+b+c} \operatorname{sech}\left(\frac{\pi}{2\sqrt{3}} \sqrt{(6a+1)^2+(6b+1)^2+(6c+1)^2}\right) = \frac{\pi}{4 \Gamma\left(\frac34\right)^4}.\tag{12}$$
UPDATE.
Paramanand Singh was able to prove the closed form for $I(4)$.
Using the substitution suggested by him, we obtain new representations for $I(8)$ and $I(12)$:
$$I(8) = \frac{2^{1/3}}{\pi^3} \int_0^1 x^{1/3}\,(1-x^2)^{-1/3}\,K(x)^2 \,dx \stackrel?= \frac23 \left(\frac{2^{2/3} \pi}{3 \Gamma\left(\frac23\right)^3}\right)^3 \tag{13}$$
$$I(12) = \frac{2}{\pi^5} \int_0^1 x\,K(x)^4\,dx = \quad ? \tag{14}$$
Here $K$ is the complete elliptic integral of the first kind.
Furthermore, in view of the definitions of Ramanujan's tau function and the Tau Dirichlet series, it follows that
$$I(24) = \int_0^{\infty} x^{10} \eta^{24}(i x) dx = \frac{10!}{(2\pi)^{11}} \sum_{n=1}^{\infty} \frac{\tau(n)}{n^{11}}. \tag{15}$$
I am not sure about that one.
ONE MORE UPDATE
With some educated guesses and Mathematica's 'RootApproximant', I have finally found a possible closed form for $I(2)$ !!!
$$I(2)=\int_0^{\infty} \eta^2(i x) dx \stackrel?= \log\left(1+ \sqrt{3}+\sqrt{3+2 \sqrt{3}}\right) \tag{16}$$
This one holds for at least 100 digits. This is surprising to me, as it's quite different from the other ones so far. This also gives me hope that a closed form for $I(5),I(7)$ etc. exists.
Best Answer
Let $q = e^{-\pi x}$ and then $$\eta(ix) = q^{1/12}\prod_{n = 1}^{\infty}(1 - q^{2n}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ and $x = K'/K$. Then $$\eta^{4}(ix) = 2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}$$ Now we can see that $x = 0$ then $k = 1$ and $x \to \infty$ implies $k \to 0$ and $$\frac{dx}{dk} = -\frac{\pi}{2kk'^{2}K^{2}}$$ and hence the integral $$I(4) = \int_{0}^{\infty}\eta^{4}(ix)\,dx = \int_{0}^{1}2^{-4/3}\left(\frac{2K}{\pi}\right)^{2}(kk')^{2/3}\cdot\frac{\pi}{2kk'^{2}K^{2}}\,dk$$ and then $$I(4) = \frac{2^{-1/3}}{\pi}\int_{0}^{1}k^{-1/3}(1 - k^{2})^{-2/3}\,dk$$ and putting $k^{2} = m$ we get $dk = \dfrac{dm}{2\sqrt{m}}$ and hence $$I(4) = \frac{2^{-4/3}}{\pi}\int_{0}^{1}m^{-2/3}(1 - m)^{-2/3}\,dm = \frac{2^{-4/3}}{\pi}\frac{\Gamma^{2}(1/3)}{\Gamma(2/3)}$$ and since $\Gamma(1/3)\Gamma(2/3) = 2\pi/\sqrt{3}$ we get the desired value of $I(4)$. I was lucky to get $I(4)$ because the $K^{2}$ term got cancelled in the evaluation of integral.
Incidentally I get the evaluation of a nice sum $$\sum_{n = 0}^{\infty} (-1)^n \dfrac{\sinh\dfrac{\pi}{\sqrt{3}}(2n + 1)}{\cosh\dfrac{\sqrt{3}\pi}{2}(2n + 1)} = \frac{\pi}{3\sqrt[3]{2}\Gamma^{3}(2/3)}$$