What is the polynomial functor for the Bag monad

The bag monad is not polynomial.

Any polynomial endofunctor must preserve pullbacks: $f^*$ and $\Pi_g$ preserve all limits since they’re right adjoints, while $\Sigma_h$, being just the forgetful functor from a slice category, is well known (and easily seen) to preserve all connected limits.

However, the bag monad doesn’t preserve pullbacks. Write $B$ for the bag monad, and consider the sets $X = \{a,b\}$, $Y = \{y,z\}$, and view their product as a pullback, $X \times Y = X \times_1 Y$. Then the canonical map $B(X \times Y) \to B(X) \times_{B(1)} B(Y)$ fails to be injective, since $\{(a,y),(b,z)\}$ and $\{(a,z),(b,y)\}$ are distinct in $B(X \times Y)$ but have the same image in $B(X)$ and $B(Y)$.

This example — and the fact that commutativity forms an obstacle to being polynomial, and similar sorts or representation — has appeared notably before in the literature, in for instance the note 3-computads do not form a presheaf category by Michael Makkai and Marek Zawadowski, and (essentially) in Carlos Simpson’s paper Homotopy types of strict 3-groupoids.

The free multiset functor is not polynomial in the standard sense; it is though in a categorified sense if you somehow keep track of the different ways two expressions are the same thanks to commutativity: for that you need to pass from the category Set to the 2-category of groupoids. See Data Types with Symmetries and Polynomial Functors over Groupoids.