Notation abuse warning: I will use the E7 series irrep names. You'll soon see why.

In the general Lie algebra, the irrep you "start" with is $J$, the
adjoint. From the series for $J\bigotimes{J}=1+J+A...$ ($1$ is the 1-dimensional irrep) you get $A$, the antisymmetric one. (The quantum dimensions up to here are all given e.g. in Westbury's "R-matrices and the magic square" paper.) The next step would be $J\bigotimes{A}=J+A+S...$ where you can "fish out" the symmetric irrep $S$ from the Clebsch Gordan series. (Does anybody have the quantum dimension $Dim(S)=S(\alpha,\beta,\gamma)$ handy, just as for $J$ and $A$? In principle I could compute it myself, but this would speed up my work considerably.)

Now comes the silly part :-) In the E7 series, $V\bigotimes{V}=1+J+S+A$ and I can now backwards compute the dimension of the defining irrep $V$ and see what comes out for, say, E8. Well, nonsense is coming out, which is hardly surprising. Still, I would have expected
integer nonsense or even better a zero ($324-273-52+1=0$ for F4 -the minus signs are only very mild cheating. :-)! Irrational nonsense is baaaad.

So: Is it possible to bring all irreps, even from different rows of the magic square, in an 1:1 correspondence (if neccessary by inventing fictive irreps which still have integer dimension) or am I completely barking up the wrong Lie?

J=E+J+A+6+7+8. This works for E7 as well as E8, and you have the closed quantum dimension formula of Vogel for J,A,6,7,8, and counting terms in the Clebsch-Gordan you know that 6 corresponds to 6, 7 to 7 and 8 to 8 (duh :-). JA=J+A+S+... Dito. You can say which irrep of E8 corresponds to S of E7. (It's a 0.) (cont.) – Hauke Reddmann Aug 24 '12 at 9:38