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 the 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*}

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*}