Suppose we are given a CNF formula $F$ with $n$ variables and $m$ clauses. The question is whether we can represent $\lnot F$ as a CNF formula with the number of clauses polynomial in $n$ and $m$?
No, not necessarily.
If $F$ is $(\neg X_1 \lor \neg Y_1) \land \dots \land (\neg X_n \lor \neg Y_n)$, then expressing $\neg F$ in CNF requires exponentially $2^n$ clauses. See, e.g., https://en.wikipedia.org/wiki/Conjunctive_normal_form#Conversion_into_CNF, Which CNF boolean formulas blow up exponentially at conversion to DNF?, and Formulas for which any equivalent CNF formula has exponential length.
However, if you allow introducing additional variables, then you can construct a formula $G$ that is equisatisfiable with $\neg F$ and such that $G$ can be represented in CNF with number of variables and clauses polynomial in $n$ and $m$.