In Trèves's Topological Vector Spaces, Distributions and Kernels, Theorem 3.1 is as follows.
A filter $\mathscr F$ on a vector space $E$ is the filter of neighborhoods of the origin in a topology compatible with the linear structure of $E$ if and only if it has the following properties:
$(3.1)\quad $ The origin belongs to every subset $U$ belonging to $\mathscr F$.
$(3.2)\quad $ To every $U \in \mathscr F$ there is $V \in \mathscr F$ such that $V + V \subset U$.
$(3.3)\quad $ For every $U \in \mathscr F$ and for every $\lambda \in \mathbb C$, $\lambda \neq 0$, we have $\lambda U \in \mathscr F$.
$(3.4)\quad $ Every $U \in \mathscr F$ is absorbing.
$(3.5)\quad $ Every $U \in \mathscr F$ contains some $V \in \mathscr F$ which is balanced.
I think condition (3.3) is redundant, i.e., can be derived from the other premises (including the fact that $\mathscr F$ is a filter). Below is a potential proof. Please let me know if it is correct or flawed.
Let $U \in \mathscr F$ and $0 \neq \lambda \in \mathbb C$. Since filters are upward closed, it suffices to find some $V \in \mathscr F$ such that $V \subset \lambda U$. Let $W \in \mathscr F$ be a balanced subset of $U$. Then $\lambda W = |\lambda| W$. If $|\lambda| \geq 1$, then $W \subseteq |\lambda| W$, so we can let $V = W$ and then we are done. Otherwise, choose $n \in \mathbb N$ such that $1 / n < |\lambda|$. Then $(1/n)W \subset |\lambda| W = \lambda W$. Iterating $(3.2)$, we can find $V \in \mathscr F$ such that $\sum_{i=1}^n V \subset W$. Then $nV \subset \sum_{i=1}^n V \subset W \subset n \lambda W$, and hence $V \subset \lambda W \subset \lambda U$. $\square$