I am not sure this answers the question, but I still wanted to provide another (geometric) viewpoint.

My issue with the other answers talking about $\text{Hom}(V,V) \cong V^* \otimes V$ is that in order to actually calculate the trace of a linear $T:V\to V$, I have to first write it in the form of something in $V^* \otimes V$. So thinking in terms of the computational process, I do have to choose a basis.

But using the geometric interpretation of trace, I don't have to do this. Suppose I have a blank piece of paper, with a single black dot on it, called $0$. Purely geometrically, I can work with the vector space structure (e.g. parallelogram rule for addition, etc.) If you further allow me to draw one ball $\frak B$ centered at $0$, and have access to the Haar measure on $V$ (the unique such assigning volume 1 to the ball I drew --- this doesn't require picking a basis) then I can calculate the trace as follows:

$$\text{tr}(T) := \lim_{t\to 0} \frac{\mu((I+tT)\frak B) - \mu(\frak B)}{t}$$

Never once did I have to prioritize a certain direction on my piece of paper; I just needed 1 dot ($0$) and 1 ball. But why is choosing a distinguished shape (ball) better than a distinguished basis?

I think the answer lies in Samra's comment about vector bundles. Consider the Mobius band as a line bundle over the circle $S^1$. Then there is no continuous choice of basis! But there is a continuous choice of ball (i.e. continuous choice of norm, which gives rise to a unit ball).

Allow me a digression to another field.

In the field of Borel combinatorics, one way of [thinking about]/[limiting] the amount of choices one makes is by having (continuum) many copies of the same object, and trying to do a process on all of them without using the axiom of choice.

For example, consider the graph with vertex set $V = \{z\in \mathbb C: |z|=1\}$ and edge set $E = \{ \{z, e^{i\alpha}z\}: z\in V\}$ and $\alpha$ an irrational multiple of $\pi$. Every connected component of this graph is a copy of (the Cayley graph w.r.t. generators $\pm 1$ of) $\mathbb Z$, so the chromatic number is 2. But (using for instance measure theory) one can show that there is no Borel (nor Lebesgue-measurable) 2-coloring $c: V\to 2$. This formalizes in some sense that in order to get a 2-coloring one must "pick a starting vertex" in each connected component (of which there are continuum many).

So this tells us that "picking a distinguished vertex in a copy of $\mathbb Z$" is not very "natural", because there is an example of a "natural" setting with infinitely many copies of $\mathbb Z$ in which there is no Borel way of picking a distinguished vertex in each copy.

Likewise, "picking a basis in a copy of $\mathbb R$" (each fiber in the Mobius bundle) is not very "natural", because there is an example of a "natural" setting with infinitely many copies of $\mathbb R$ in which there is no continuous way of picking a distinguished basis vector in each copy. However, there is a continuous way of picking a distinguished ball in each copy.

Hopefully this communicates some sense in which "picking a basis" is not natural, and how one can avoid this "non-naturality" but still compute the trace.

(Maybe in fact one can define a version of the trace for all $\sigma$-compact locally compact abelian groups --- which admit Haar measure, and an equivariant distance function leading to a compact unit ball, I think)