Special values of the modular J invariant A special value:
$$
J\big(i\sqrt{6}\;\big) = \frac{(14+9\sqrt{2}\;)^3\;(2-\sqrt{2}\;)}{4}
\tag{1}$$
I wrote $J(\tau) = j(\tau)/1728$.  
How up-to-date is the Wikipedia listing of known special values for the modular
j invariant ?
Value (1) is not on it.
Alternatively, is there a compilation of special values elsewhere?   
The value (1) is related to this hypergeometric value, too:
$$
{}_2F_1\left(\frac{1}{6},\frac{1}{3};1;\frac{1}{2}\right) =
\eta(i\sqrt{6}\;)^2 \,2^{1/2} \,3^{3/4} \,(1+\sqrt{2}\;)^{1/6}
\tag{2}$$
Here, $\eta(\tau)$ is the Dedekind eta function 
(I came to this while working on my unanswered question
at math.SE)
 A: 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*}
A: The value of $\eta(i\sqrt{6})$ and $\eta(i\sqrt{3/2})$ involves the use of gamma function values on a 24 basis, so we have:
$$\eta(i\sqrt{6})=\frac{1}{2^{3/2}3^{1/4}}\big(\sqrt{2}-1\big)^{1/12}\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}$$
$$\eta(i\sqrt{3/2})=\frac{1}{2^{5/4}3^{1/4}}\big(\sqrt{2}+1\big)^{1/12}\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}$$.
For completeness, we have:
$$\eta(i\sqrt{2/3})=\frac{1}{2^{3/2}}\big(\sqrt{2}+1\big)^{1/12}\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}$$.
[added GEdgar] also
$$
\eta(i\sqrt{1/6})=
\frac{1}{2^{5/2}}\big(\sqrt{2}-1\big)^{1/12}
\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}
$$
A: You are at $z=i\sqrt{6}$. Hence you can use singular modulus $k_r$. It holds in general (when $z=i\sqrt{r}$)
$$
j_r=\frac{256(k_r^2+(k'_r)^4)^3}{(k_rk'_r)^4}\textrm{, }k'_r=\sqrt{1-k_r^2}\textrm{, }\forall r>0.
$$
But $k_6=\lambda^*(6)=(2-\sqrt{3})(\sqrt{3}-\sqrt{2})$ (see here). Hence you get your value.
A: I don't think that it's too hard, in principle, to compute many such examples. The Wikipedia article lists a bunch, but it's not clear why they chose those particular ones. The point, of course, is that $j(\tau)$ with $\tau$ imaginary quadratic generates a proper ideal in an order in a ring class field associated to $\tau$ and $\mathbb Q(\tau)$. As long as this field isn't too large, a computer algebra system will get you generators for the ideal, and thence express $j(\tau)$ in terms of the generators. 
I'll also note in passing that your $J(i\sqrt 6)$ is actually in the ring of integers $\mathbb Z[\sqrt2]$ of $\mathbb Q(\sqrt2)$, since it is equal to
$$ J(i\sqrt6) = (9+7\sqrt2)^3(-1+\sqrt2). $$
Not sure if that's helpful for the application you have in mind.
