Product norm on infinite product space

Proposition. If $X_i, i \in \mathbb{N}$ is any sequence of nonzero topological vector spaces, then the product topology on $X = \prod_i X_i$ is not normable.

Proof. Suppose to the contrary there is a norm $\|\cdot\|$ on $X$ which induces the product topology. Let $B \subset X$ be the open unit ball of $\|\cdot\|$. $B$ is open in the product topology and contains 0, so we can find a basic open neighborhood of 0 which is contained in $B$. That is, there is a set of the form $U = U_1 \times U_2 \times \dots \times U_n \times X_{n+1} \times \cdots$, where $U_i \subset X_i$ is open and nonempty, such that $0 \in U \subset B$. Choose any nonzero $x_{n+1} \in X_{n+1}$ and set $x = (0,\dots, 0, x_{n+1}, 0, \dots)$. Then for any $t \in \mathbb{R}$ we have $tx \in U \subset B$. If we take $t = 2/\|x\|$, we have $\|tx\| = 2$ which is absurd since $B$ is the unit ball.

In short, the problem is that every open set of the product topology contains a line, and balls of a norm do not.

The correct question is not "can I put a norm on the ordinary product?" but "what should the product of infinitely many Banach spaces be?" and then one asks what its relation is to the ordinary product. One answer is to take the subspace of $V^{\mathbb{N}}$ consisting of all sequences $v_i$ such that $$\sum_{i=1}^{\infty} \| v_i \|$$

converges, with the above as norm. This is the coproduct of countably many copies of $V$ in the category $\text{Ban}_1$ of Banach spaces and weak contractions. Another is to take the subspace of $V^{\mathbb{N}}$ consisting of all sequences $v_i$ such that $$\sup_i \| v_i \|$$

exists, with the above as norm. This is the product of countably many copies of $V$ in $\text{Ban}_1$.

It follows that the forgetful functor $\text{Ban}_1 \to \text{Set}$ sending a Banach space to its set of vectors preserves neither products nor coproducts. It is better to use another forgetful functor $\text{Hom}(1, -)$ sending a Banach space to its unit ball, which preserves products (in fact since it is representable it preserves all limits).

The topology on $V^{\mathbb{N}}$ is generated by the Fréchet metric $$d((x_j),(y_j)) = \sum_{k=0}^{\infty} \frac{|x_k - y_k|}{2^k(1 + |x_k - y_k|)}$$ which seems to be in the spirit of what you suggested. This isn't induced by any norm, though.