Baryons in flavor $SU(N)$ (in ChPT) For flavor $SU(2)$ (Isospin) we have two $\frac{1}{2}^+$ baryons, the nucleons. For flavor $SU(3)$ we have the eight baryons in the octet. In a world with $N$ light quarks we would see a baryon multiplet of dimension $\frac{N}{3}(N^2-1)$.
Such a theory would see the chiral symmetry breaking $SU(N)\times SU(N) \to SU(N)$ creating $N^2-1$ Goldstone bosons $\phi^a, a=1,\dots,N^2-1$. These mesons are usually parametrized in the $N\times N$ matrix $U=\exp\left( i T^a \phi^a \right)$, where $T^a$ are the generators of $SU(N)$. These fields than transform under $(L,R)\in SU(N)\times SU(N)$ as $U \mapsto R U L^\dagger$.
So far so good (please correct me if I already made any mistakes, I think this is correct). My question now is:
How can we include the $\frac{N}{3}(N^2-1)$ baryons in a Lagrangian?
We need to find a parametrization for these baryon fields and we have to find out how they transform under $SU(N) \times SU(N)$.
It feels like we got a bit lucky in reality, since for $N=2$: $\frac{N}{3}(N^2-1)=N=2$, and for $N=3$: $\frac{N}{3}(N^2-1)=N^2-1=8$ is the number of generators of $SU(3)$. So in these cases we can use the isospin doublet $N = \begin{pmatrix} p\\n\end{pmatrix}$ and for octet baryons we can use $B=\sum_{a=1}^{8} \frac{B^{a} \lambda^{a}}{\sqrt{2}}=\left[\begin{array}{ccc}{\Sigma^{0} / \sqrt{2}+\Lambda / \sqrt{6}} & {\Sigma^{+}} & {p} \\ {\Sigma^{-}} & {-\Sigma^{0} / \sqrt{2}+\Lambda / \sqrt{6}} & {n} \\ {\Xi^{-}} & {\Xi^{0}} & {-\sqrt{\frac{2}{3}} \Lambda}\end{array}\right].$
Is there literature for chiral perturbation theory for $N$ light flavors? How would one go about including baryons?
 A: Baryons in ChPT are an advanced subject, so I won't presume to do your Googling for you. But you recall baryons are fermions, so you don't need gimmicks: $SU(N)\times SU(N)$ is realized linearly on vectors of an m-dimensional representation. 
Recall the nucleon isodoublet 
$$\begin{pmatrix} p\\n\end{pmatrix}$$ is acted upon on the left by both the L and the R chiral operators,
$$
L^i=\tfrac{1}{2} \tau^i P_L, \qquad R^i=\tfrac{1}{2} \tau^i P_R,
$$
where the two factor groups commute by virtue of the chiral projectors,
the flavor group (algebra) being just the vector, $V^i=\tfrac{1}{2} \tau^i $.
So, how does the group act on the Δ isoquartet $(Δ^{++},Δ^+, Δ^0, Δ^-)^T$?  Same way, except you utilize the quartet isospin generators, mutatis mutandis...
Same for SU(3); you could, if you wished, act on the left of the octet 8-vector 
$$ B^a=
\begin{pmatrix}\sqrt{2}(\Sigma^++\Sigma^-)\\   
i\sqrt{2}(\Sigma^+ -\Sigma^-)\\
 \Sigma^0 \\
\sqrt{2}( p+\Xi^-)\\
i\sqrt{2}( p-\Xi^-)\\
\sqrt{2}( n+\Xi^0)\\
i\sqrt{2}( n-\Xi^0)\\ \Lambda
\end{pmatrix}
$$
by the adjoint rep (generators are the 8 structure constant matrices) instead of your matrix realization, now with the chiral gamma matrix projectors tacked on. (One assumes you have done the exercise of linking the two!) 
But that would also suggest to you how to deal with the baryon decuplet, one row of which we just did above! (However,  you'd have to chase down the 10-dim generator matrices.)
Proceed to SU(4), where the parents of the octet and the decuplet above are both 20 s, by coincidence. 
For generic flavor N, the mixed symmetry octet blossoms to $N(N^2-1)/3$-tuplet reps; but the symmetric decuplet to $N(N+1)(N+2)/6$-tuplets, etc.
