$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$
After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:
$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$
Next, substituting $t=\frac{1}{u}$ yields:
$$\begin{align}
I
&=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\
&=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\
&=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\
\implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}.
\end{align}$$
Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:
$$\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\
&=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\
&=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\
&=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\
&=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\
&=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}
\end{align}$$
Hence,
$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$
Possible Alternative: You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.
$$\begin{align}
\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\
&=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\
&=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\
&=2\,K{(-1)}\\
&=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\
&=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}.
\end{align}$$
The conjecture is true, as are the other cases reported in the comments
where $f(z) := {}_2F_1 \left( \frac13, \frac13; \frac56; z \right)$
takes algebraic values for special rational values of $z$.
There are a few others obtained from the symmetry $z \leftrightarrow 1-z$
(these ${}_2F_1$ parameters correspond to a hyperbolic triangle group
with index $6,6,\infty$ at $c=0,1,\infty$, so the $z=0$ and $z=1$
indices coincide); e.g. $f(-1/3) = 2 / 3^{2/3}$ pairs with
$f(4/3) = 3^{-2/3} (5-\sqrt{-3})/2$. ($z=1/2$ pairs with itself,
and the pair $f(-4)$ and $f(5)$ has been noted already;
the OP's $f(-27) = -4/7$ pairs with $f(28) = \frac12 - \frac3{14} \sqrt{-3}$.)
Somewhat more exotic are
$$
f\big({-}4\sqrt{13}\,(4+\sqrt{13})^3\big) = \frac7{13\,U_{13}}\\
f\big({-}\sqrt{11}\,(U_{33})^{3/2}\big) = \frac{6}{11^{11/12}\, U_{33}^{1/4}},
$$
with fundamental units $U_{13}=\frac{3+\sqrt{13}}2,\;U_{33}=23+4\sqrt{33}$ and further values at algebraic conjugates and images under
$z \leftrightarrow 1-z$.
In general, for $z<1$ the integral formula for $f(z)$ relates it with
$$
\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }
$$
which is half of a "complete real period" for the holomorphic differential
$dx/y$ on the curve $C_z : y^6 = (1-x)^3 x^4 (1-zx)^2$. This curve
has genus $2$, but is in the special family of genus-$2$ curves
with an automorphism of order $3$ (multiply $y$ by a cube root of unity),
for which both real periods are multiples of the real period of a
single elliptic curve $E_z$ (a.k.a. a complete elliptic integral).
In general the resulting formula doesn't simplify further, but when
$E_z$ has CM (complex multiplication) its periods can be expressed
in terms of gamma functions. For $z = -27$ and the other special values
listed above, not only does $E_z$ have CM but the CM ring is contained in
${\bf Z}[\rho]$ where $\rho = e^{2\pi i/3} = (-1+\sqrt{-3})/2$.
Then the $\Gamma$ and $\pi$ factors of the period of $E_z$ exactly
match those in the integral formula, leaving us with
an algebraic value of $f(z)$. It turns out that the choice $z = -27$
makes $E_z$ a curve with complex multiplication by ${\bf Z}[7\rho]$.
The others from the comments lead to ${\bf Z}[m\rho]$ with $m=1,2,3,5$,
and the examples where $z$ is a quadratic irrationality come from
${\bf Z}[13\rho]$ and ${\bf Z}[11\rho]$.
One way to get from $C_z$ to $E_z$ is to start from the change of variable
$u^3 = (1+cx)/x$, which gives
$$
f(z) = \int_{\root 3 \of {1-z}}^\infty \frac{3u \, du}{\sqrt{(u^3+z)(u^3+z-1)}}.
$$
and identifies $C_z$ with the hyperelliptic curve $v^2 = (u^3+z)(u^3+z-1)$.
Now in general a curve $v^2 = u^6+Au^3+B^6$ has an involution $\iota$ taking
$u$ to $B^2/u$, and the quotient by $\iota$ is an elliptic curve;
we compute that this curve has $j$-invariant
$$
j = 6912 \frac{(5+2r)^3}{(2-r)^3(2+r)}
$$
where $A = rB^3$. (There are two choices of $\iota$, related by
$v \leftrightarrow -v$, and thus two choices of $j$, related by
$r \leftrightarrow -r$; but the corresponding elliptic curves
are $3$-isogenous, so their periods are proportional.)
In our case $r = A/B^3 = -(2z+1)/\sqrt{z^2+z}$ (in which the
$z \leftrightarrow 1-z$ symmetry takes $r$ to $-r$). Taking $z=-27$
yields $j = -2^{15} 3^4 5^3 (52518123 \pm 11460394\sqrt{21})$,
which are the $j$-invariants of the ${\bf Z}[7\rho]$ curves;
working backwards from the $j$-invariants of the other
${\bf Z}[m\rho]$ curves we find the additional values of $z$
noted in the comments and earlier in this answer.
Best Answer
I would like to thank M.N.C.E. for suggesting the use of the identity $$2\log(x)\log(y) = \log^{2}(x) + \log^{2}(y) - \log^{2} \left(\frac{x}{y} \right), $$ where $x$ and $y$ are positive real values.
As I stated here, we have
\begin{align} &\int_{0}^{\infty} [\operatorname{Ei}(-x)]^{4} \, dx \\ &= -4 \int_{0}^{\infty} [\operatorname{Ei}(-x)]^{3} e^{-x} \, dx \tag{1} \\ &= 4 \int_{0}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} e^{-(w+y+z+1)x} \, dw \, dy \, dz \,dx \tag{2} \\& =4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \int_{0}^{\infty} e^{-(w+y+z+1)x} \, dx \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \frac{1}{w+y+z+1} \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \left(\frac{1}{w} - \frac{1}{w+y+z+1} \right) \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \log(y+z+2) \, dy \, dz \\ &= 8 \int_{1}^{\infty} \int_{1}^{z} \frac{1}{yz} \frac{1}{y+z+1} \log(y+z+2) \, dy \, dz \tag{3} \\ &= 8 \int_{2}^{\infty} \int_{u-1}^{u^{2}/4} \frac{1}{v} \frac{1}{u+1} \log(u+2) \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{4} \\& =16 \int_{2}^{\infty} \frac{\log(u+2)}{u+1} \int_{0}^{u-2} \frac{1}{u^{2}-t^{2}} \, dt \, du \tag{5} \\& =16 \int_{2}^{\infty} \frac{\log(u+2)}{u+1} \frac{1}{u} \, \operatorname{artanh} \left( \frac{u-2}{2}\right) \, du \\ &= 8 \int_{2}^{\infty} \frac{\log(u+2) \log(u-1)}{u(u+1)} \, du \\ &= 8 \Bigg(\int_{0}^{1/2} \frac{\log(1-u)\log(1+2u)}{1+u} \, du - \int_{0}^{1/2} \frac{\log(u) \log(1+2u)}{1+u} \, du \\ &- \int_{0}^{1/2} \frac{\log(u) \log (1-u) }{1+u} \, du + \int_{0}^{1/2} \frac{\log^{2}(u)}{1+u} \, du \Bigg). \tag{6} \end{align}
$(1)$ Integrate by parts.
$(2)$ $\operatorname{Ei}(-x) = - \int_{x}^{\infty} \frac{e^{-t}}{t} \, dt = - \int_{1}^{\infty} \frac{e^{-xu}}{u} \, \mathrm du $
$(3)$ The integrand is symmetric about the line $z=y$.
$(4)$ Make the change of variables $u= y+z$, $v=yz$.
$(5)$ Make the substitution $ t^{2}=u^2-4v$.
$(6)$ Replace $u$ with $\frac{1}{u}$.
Then using the identity I mentioned at the beginning of this answer, we have
$$ \begin{align} &\int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx \\ &= -4\int_{0}^{1/2} \frac{\log^{2} \left(\frac{1-u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1-u} \right)}{1+u} \, du \\& = -12 \int_{1/4}^{1} \frac{\log^{2} (w)}{(1+2w)(2+w)} \, dw +4 \int_{0}^{1/4} \frac{\log^{2} (y)}{(1-2y)(1-y)} \, dy + 4 \int_{0}^{1} \frac{\log^{2}(z)}{(1+2z)(1+z)} \, dz \\&= -6 \int_{1/4}^{1} \frac{\log^{2} (w)}{(1+2w)(1+\frac{w}{2})} \, dw +4 \int_{0}^{1/4} \frac{\log^{2} (y)}{(1-2y)(1-y)} \, dy + 4 \int_{0}^{1} \frac{\log^{2}(z)}{(1+2z)(1+z)} \, dz. \end{align}$$
EDIT:
Partial fraction decomposition followed by two applications of integration by parts shows that $ \begin{align} \int \frac{\log^{2}(x)}{(1+ax)(1+bx)} \, dx = \frac{1}{a-b} \Big(&2\big(\operatorname{Li}_{3}(-bx) - \operatorname{Li}_{3}(-ax) \big) + 2 \log (x) \big( \operatorname{Li}_{2}(-ax) - \operatorname{Li}_{2}(-bx)\big) \\ &+\log^{2}(x) \big(\ln(1+ax) - \log(1+bx) \big) \Big) + C. \end{align}$
This primitive can be used to determine the value of all three definite integrals above.
The final result won't immediately be in the form given by Cleo. Getting it in that form will require the use of polylogarithm identities.