Let $F$ be a field. Let $V$ be an $n$-dimensional vector space over $F$. For concreteness, take $n$ to be 3.
I know that $V$ is isomorphic to $F^3$. I think that's what allows us to think of the elements of $V$ as $3$-tuples with entries in $F$. (Is this true?)
Now, $V$ has no canonical basis.
Yet, in $F^3$ the elements $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ seem "special". (If anything, because $0$ and $1$ are the two most important elements in any field...)
Still, $3$-tuples (by themselves) don't have meaning in $V$, because a $3$-tuple like $(1,0,0)$ can represent any vector in $V$: without a basis, a $3$-tuple with entries in $F$ can't unambiguously define an element/vector of $V$, hence "it isn't well-defined".
So V and $F^3$ seem "similar", but not "exactly alike".
- Is $V$ canonically isomorphic to $F^3$? Why (not)?
- Can something be said about the space of all isomorphisms between $V$ and $F^3$?
- If $V$ isn't canonically isomorphic to $F^3$, then isn't it "wrong" to think about vectors in $V$ as $3$-tuples over $F$?
As a matter of pedagogical preference, I personally don't like introducing vector spaces through $F^n$, largely because of this canonical basis. I find that the idea of one basis being more important/natural than another basis largely impedes the core understanding of finite-dimensional linear algebra.
The fact is, there are often many (infinitely many, when $F$ is infinite) bases for a (non-trivial) finite-dimensional vector space, and all of them are just as important as each other. We tend to have aesthetic preferences regarding certain bases over others (in $F^n$ or in spaces of polynomials, etc), but from an algebraic standpoint, they're all fungible.
I prefer to address $F^n$ when it comes time to talk about coordinate vectors. I prefer to leave $F^n$ essentially to be the space of coordinate vectors of an abstract $n$-dimensional space $V$, under a fixed basis. I want students to see a vector like $(1, 2, 3)$ and translate it into a linear combination $v_1 + 2v_2 + 3v_3$! Then $(1, 0, 0)$ is "special" because it's the coordinate vector for your first basis vector $v_1$. That is to say, it's not special in any kind of absolute sense, but it's a really "natural" vector in the context of coordinates for our fixed basis $(v_1, v_2, v_3)$.
That's what I think of when I see $F^n$: they're just coordinate vectors for some basis. The space $F^n$ is no more than the agreed meeting point connecting to all $n$-dimensional vector spaces, via each of their many bases.
So, to more directly answer your question, "no" an abstract finite-dimensional space does not have a particular basis that is more "special" than the others. Choices of basis are often made for convenience or cosmetic purposes.
To answer your second question, the isomorphisms from $V$ to $\Bbb{F}^3$ take the form $ST_0$, where $T_0 : V \to \Bbb{F}^3$ is some fixed isomorphism, and $S$ ranges over all isomorphisms from $\Bbb{F}^3$ to $\Bbb{F}^3$ (which can be represented by the group of invertible $n \times n$ matrices, even canonically so). However, this relies on an arbitrary fixed choice of $T_0 : V \to \Bbb{F}^3$, which is in essence, an arbitrary choice of basis $(T_0^{-1}(e_1), \ldots, T_0^{-1}(e_n))$. So, again, structurally we have something familiar (invertible matrices), but the actual correspondences between the structures are not "canonical".
As for your third question, I would say it is "wrong" to think of the vectors as $n$-tuples. Coordinate vectors should be a tool to help you solve problems, not the complete underpinning of your understanding. Like any mathematical tool, there are situations where it will apply, and situations where it won't, and having a broader understanding helps you differentiate between the two situations. In particular, you should not be lost when it comes to infinite-dimensional spaces (e.g. the space of real functions).