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The answer is no: the projective tensor product is not left-exact on $\mathrm{Ban}_1$.

There are several confusions in the question, that the following three points should hopefully clarify:

1. For usual Banach spaces (that is over real or complex numbers), the functor $\cdot \otimes_\pi V$ is in general not left-exact in any way, see below for details.
2. Some number theorists have had the strange idea to redefine "Banach space" as "Banach space over a non-archimedean local field with ultrametric norm". It is not surprising that this can cause confusion, and your question is a perfect illustration of that: the reference you give only deals with such ultrametric spaces, and is therefore irrelevant for your question.
3. If $0\to X\to Y \to Z\to 0$ is a short exact sequence of vector spaces where all spaces are Banach spaces and all maps are continuous, then by the open mapping theorem, $X\to Y$ is an isomorphism on its image and $Y \to Z$ is open, so $X$ and $Z$ can be given equivalent norms so that $X\to Y$ becomes an isometry and $Y\to Z$ becomes the quotient map $Y \to Y/X$. In other words, every short exact sequence in $\mathrm{Ban}$ and in the loose sense is isomorphic to short exact sequence in $\mathrm{Ban}_1$ in your "much more restrictive" sense. In particular, for a functor on $\mathrm{Ban}$ that happens to preserve $\mathrm{Ban}_1$, all the notions of left-exactness that you discuss in your question are formally equivalent.

As promised, here is the explanation for 1: there are Banach spaces $X,Y,V$ where $X$ is a closed subspace of $Y$ but the map $X \otimes_\pi V \to Y \otimes_\pi V$ is not injective.

For example, take an inclusion $X \subset Y$ where $Y$ has the Approximation property (AP) but not $X$. The only explicit example I can give today is $X=B(\ell_2)$ and $Y = \ell_\infty(B_{X^*}(0,1))$ where as usual, $X$ is regarded as a subspace of the bounded functions on the unit ball of its dual by evaluation. But there are many other examples, much harder to define, see the wikipedia page. The reason for this choice is that a classical theorem by Grothendieck says that the map $Z\otimes_\pi V \to Z\otimes_\varepsilon V$ is injective (where $\otimes_\varepsilon$ is the injective tensor product) for every $V$ if and only if $Z$ as the AP, see this answer. So pick $V$ such that $X\otimes_\pi V \to X\otimes_\varepsilon V$ is not injective, for example $V=X^*$. Consider the following commutative diagram $$\require{AMScd} \begin{CD} X\otimes_\pi V @>{1}>> X\otimes_\varepsilon V\\ @VV{2}V @VV{3}V \\ Y\otimes_\pi V @>{4}>> Y\otimes_\varepsilon V \end{CD}.$$ We have seen that $1$ is not injective, whereas $4$ is. Therefore, $2$ is not injective, QED.