There are $2^n$ maximal 3CNFs; here I am assuming that each 3-clause contains at most three different literals (corresponding to different variables), and that the clauses are different and unordered (both within each clause, and across the formula). Indeed, I show below that if $\varphi$ is a maximal 3CNF then there is a truth assignment $\alpha$ such that $\varphi$ consists of all 3-clauses consistent with $\varphi$, and conversely, each such formula is maximal.
Indeed, let $\varphi$ be a maximal 3CNF. Suppose first that $\varphi$ has at least two different truth assignments $\alpha,\beta$. There is a variable $x_i$ whose truth value is different in $\alpha$ and $\beta$, say $x_i$ is true in $\alpha$ but false in $\beta$. In particular, the clause $x_i$ is not in $\varphi$ (since $\beta$ doesn't satisfy it), and if we add it then the formula remains satisfiable. We conclude that $\varphi$ has a unique satisfying assignment $\alpha$. If $\varphi$ doesn't contain all 3-clauses consistent with $\alpha$ then we can add the missing ones while keeping the formula satisfiable. Thus $\varphi$ must consist of all 3-clauses consistent with $\alpha$.
Conversely, suppose that $\varphi$ consists of all 3-clauses consistent with $\alpha$. Since $\varphi$ contains all clauses of width 1, $\alpha$ is the unique satisfying assignment. By definition, any other 3-clause is inconsistent with $\alpha$, and so $\varphi$ is maximal.
You might be interested in E3CNFs, in which each 3-clause contain exactly three literals, each corresponding to a different variables. These can probably be analyzed using the same approach - if you're interested, you're welcome to work it through.