I do not know enough about CM (complex multiplication?) or
class field theory to tell whether Joe's answer is
sensible or feasible. And (so far) we have received no
references as I had hoped. So eventually I came up with
a proof, perhaps more elementary.
Write $\tau_1 = i\sqrt{6}/6$. Then
$$
\frac{-1}{\tau_1} = i\sqrt{6} = 6\tau_1 .
$$
Now both $j(\tau)$ and $j(6\tau)$ are
modular functions for the group $\Gamma_0(6)$.
Thus, they are algebraically related to each other.
And since
$$
j(6\tau_1) = j\left(\frac{-1}{\tau_1}\right) = j(\tau_1),
$$
if we put these into the algebraic relation we get
an algebraic equation for $j(\tau_1)$.
Evaluating $j(\tau_1)$ numerically, we can tell which zero of
that equation is the right one.
To determine the algebraic relationship between
$j(\tau)$ and $j(6\tau)$, I chose the Hauptmodul
$$
j_{6E}(\tau) = \frac{\eta(2\tau)^3\eta(3\tau)^9}
{\eta(\tau)^3\eta(6\tau)^9}
= q^{-1} + 3 + 6q + 4q^2 - 3 q^3 +\dots
$$
Then $j(\tau)$ and $j(6\tau)$ are both rational functions of it.
My computer calculates—writing $x=j_{6E}(\tau)$:
\begin{align*}
j(\tau) &= \frac{(x+4)^3(x^3+228 x^2 + 48 x + 64)^3}
{x^2(x-8)^6(x+1)^3}
\\
j(6\tau) &=\frac{x^6(8/x^3+12/x^2+6/x-1)^3(1-2/x)^3}
{8/x^3+15/x^2+6/x-1}
\end{align*}
When $\tau=\tau_1$, these are equal. Equating them,
we get an equation to solve for $x$.
Of degree $10$. (It factors somewhat.)
Maple doesn't numerically evaluate eta functions directly.
But we can write them in terms of theta functions
$$
j_{6E}(\tau) =
{\frac {{{\rm e}^{-2\,i\pi \,\tau}}
\theta_4 \left( \pi \,\tau,{{\rm e}^{6\,i\pi \,\tau}}
\right) ^{3}
\theta_4 \left(
\frac{3}{2}\,\pi \,\tau,{{\rm e}^{9\,i\pi \,\tau}}
\right) ^{9}}{ \theta_4 \left( \frac{1}{2}\,\pi \,
\tau,{{\rm e}^{3\,i\pi \,\tau}} \right) ^{3}
\theta_4 \left( 3\,\pi \,\tau,{{\rm e}^{18\,i\pi \,\tau}}
\right) ^{9}}}
$$
which Maple does numerically evaluate.
We get
$$
j_{6E}(\tau_1) \approx 16.48528137423857 .
$$
The only zero of our polynomial that matches this is
$x = 8 + 6\sqrt{2}$. Plugging it in, we get
\begin{align*}
j(i\sqrt{6}\,) &= 2417472+1707264\sqrt{2}
=2^6 \;3^3 \;(1+\sqrt{2}\,)^5\;(3\sqrt{2}-1)^3
\\
J(i\sqrt{6}\,) &= 1399+988\sqrt{2}=
(1+\sqrt{2}\,)^5\;(3\sqrt{2}-1)^3 .
\end{align*}
