This post may be coincide with some of the contents here.
From Conway, A course in functional analysis, page 104.
If $H$ is a finite dimensional vector space and $F_{1},F_{2}$ are two topologies on $H$, that makes $H$ into a TVS. Then $F_{1}=F_{2}$.
I do not really know how to prove this because $H$ may not be normable if all its non-empty open sets are unbounded. If $H$ is normable, then a subtle argument showed any norm on a finite dimensional space is equivalent to each other, hence $F_{1}=F_{2}$. (See Conway page 69). But I do not really know what to do when $H$ is just a finite dimensional TVS with no additional structure given. Even in the one dimensional case, it seems to me that $H$ may have different topologies. However, I also do not know how to construct one which is continuous with respect to addition and multiplication but not coincide withthe usual topology (in one dimensional case).
With any bijection between two sets, if one of the sets carries a topology, you can transport the topology to the other. The bijection then becomes a homeomorphism.
For topological vector spaces, of course not every bijection will transport a TVS topology on the one space to a TVS topology on the other, but a linear isomorphism will (elementary but tedious verifications left as an exercise).
So given any finite-dimensional $\mathbb{K}$ vector space $V$, with any linear isomorphism, we can transport its topology to $\mathbb{K}^n$.
The claim is: If the topology $\mathcal{T}$ on $V$ was a Hausdorff TVS topology, then the topology $\mathcal{T}_V$ on $\mathbb{K}^n$ induced by the linear isomorphism is the standard (product) topology $\mathcal{S}_n$ on $\mathbb{K}^n$.
Let $(e_i)_{1 \leqslant i \leqslant n}$ the standard basis in $\mathbb{K}^n$.
First, we note that the map $\lambda \mapsto \lambda\cdot e_i$ is continuous as a map $\bigl(\mathbb{K},\,\mathcal{S}_1\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)$, since $\mathcal{T}_V$ is a TVS topology.
Hence the map $\mathbb{K}^n \to \bigl(\mathbb{K}^n\bigr)^n$ given by $(\lambda_1,\, \dotsc,\,\lambda_n) \mapsto (\lambda_1\cdot e_1,\, \dotsc,\, \lambda_n\cdot e_n)$ is continuous for the product topologies obtained from $\mathcal{S}_1$ resp. $\mathcal{T}_V$.
Now, $\mathcal{T}_V$ being a TVS topology, addition is continuous, therefore
$$\begin{gather} \Phi \colon \bigl(\mathbb{K}^n,\, \mathcal{S}_n\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)\\ (\lambda_1,\, \dotsc,\, \lambda_n) \mapsto \sum_{i = 1}^n \lambda_i \cdot e_i = (\lambda_1,\, \dotsc,\, \lambda_n) \end{gather}$$
is continuous ($\mathcal{S}_n$ is the product topology of $\mathcal{S}_1$).
Therefore:
The standard topology is the finest TVS topology that $\mathbb{K}^n$ can carry.
Note that we did not assume the Hausdorff property, so that result holds even if Hausdorffness is not required.
Next, we remark
In a topological $\mathbb{K}$ vector space $E$, $0$ has a neighbourhood basis consisting of balanced sets.
(A set $B \subset E$ is balanced if $(\forall \lambda \in \mathbb{K})(\lvert \lambda\rvert \leqslant 1 \Rightarrow \lambda\cdot B \subset B)$.)
That follows from the continuity of scalar multiplication $\mathbb{K} \times E \to E$ in $(0,\,0)$, since that demands that for each neighbourhood $N$ of $0$ in $E$, there is a neighbourhood $D\times M$ of $(0,\,0)$ such that $D\cdot M \subset N$. $D$ contains a disk $D_\varepsilon = \{z \colon \lvert z\rvert < \varepsilon\}$ and $D_\varepsilon \cdot M$ is easily verified to be balanced.
Now, let's finish the proof.
If $\mathcal{T}$ is Hausdorff, so is $\mathcal{T}_V$ (trivial verification).
The closed unit ball $B$ in $\mathbb{K}^n$ is compact in $\mathcal{S}_n$, and so is its boundary $S$. Since $\Phi$ is continuous, and $\mathcal{T}_V$ Hausdorff, $\Phi(B)$ and $\Phi(S)$ are compact (in $\mathcal{T}_V$).
Since $0 \notin \Phi(S)$, for each $x \in \Phi(S)$, there is a balanced neighbourhood $U_x$ of $0$ and a neighbourhood $W_x$ of $x$ such that $U_x \cap W_x = \varnothing$.
Since $\Phi(S)$ is compact, there exist finitely many $x_1,\, \dotsc,\, x_k$ such that $\Phi(S) \subset \bigcup\limits_{i = 1}^k W_{x_i}$. Then $U := \bigcap\limits_{i=1}^k U_{x_i}$ is a balanced ($\mathcal{T}_V$) neighbourhood of $0$ that doesn't intersect $\Phi(S)$.
$\Phi^{-1}(U)$ is a balanced ($\mathcal{S}_n$) neighbourhood of $0$ that doesn't intersect $S$, hence $\Phi^{-1}(U) \subset \overset{\circ}{B}$, whence $U \subset \Phi(\overset{\circ}{B})$.
By linearity, it follows that for every $\mathcal{S}_n$-neighbourhood $N$ of $0$, $\Phi(N)$ is a $\mathcal{T}_V$-neighbourhood of $0$.
That in turn implies that $\Phi$ is open.
An open and continuous bijection is a homeomorphism, hence $\mathcal{T}_V = \mathcal{S}_n$. $\qquad$ c.q.f.d.