Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.
As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent. In fact, this argument works for regular maps $f\colon V\to\mathbb{C}^n$ for any variety $V$. There is no need to restrict the domain to be $\mathbb{C}^n$.
