An object $V$ in a symmetric monoidal category is said to be dualizable with dual $V^{\ast}$ if you can find maps

$$\text{ev} : V^{\ast} \otimes V \to 1$$

and

$$\text{coev} : 1 \to V \otimes V^{\ast}$$

such that the zigzag identities hold (the same zigzag identities in the unit-counit definition of an adjunction). It's a formal exercise to show that if $V$ is dualizable with dual $V^{\ast}$ then $V^{\ast}$ is dualizable with dual $V$; moreover, if the ambient symmetric monoidal category is closed, then you can always take $V^{\ast} = [V, 1]$ is necessarily the internal hom of $V$ and the unit, and $\text{ev}$ to be the evaluation map. From here it's not hard to show that

$$V \to [[V, 1], 1] \cong (V^{\ast})^{\ast}$$

is an isomorphism for any dualizable $V$; in other words, dualizable objects are reflexive. It's also true that if $V$ is dualizable then

$$[V, W] \cong V^{\ast} \otimes W$$

for any $W$. None of this requires making any choices and is completely canonical once you show that if duals (together with their evaluation and coevaluation maps) exist then they are unique up to unique isomorphism.

Now, in the case of vector spaces, the dualizable objects are precisely the finite-dimensional ones. Whether you need to exhibit a basis to prove this depends on whether your definition of finite-dimensional is "has a basis." Here it is the coevaluation map which it is troublesome to define without a choice of basis; with a choice of basis $e_i$ it's given by

$$\text{coev} : 1 \ni 1 \mapsto \sum e_i \otimes e_i^{\ast} \in V \otimes V^{\ast}$$

where $e_i^{\ast}$ is the dual basis. But this does not weaken the naturality of anything that's been said here, again because of the important fact that duals (together with their evaluation and coevaluation maps) are unique if they exist.

As in Oscar Cunningham's answer, if your definition of "finite-dimensional" happens to be "the natural map $V \otimes V^{\ast} \to [V, V]$ is an isomorphism" (this map always exists and requires nothing to define) then you can define the coevaluation map to be the inclusion of the identity element into $[V, V]$. At some point you will want to show that something is finite-dimensional, though, and then you might need to pick a basis. But that's okay.

An object $V$ in a symmetric monoidal category is said to be dualizable with dual $V^{\ast}$ if you can find maps

$$\text{ev} : V^{\ast} \otimes V \to 1$$

and

$$\text{coev} : 1 \to V \otimes V^{\ast}$$

such that the zigzag identities hold (the same zigzag identities in the unit-counit definition of an adjunction). It's a formal exercise to show that if $V$ is dualizable with dual $V^{\ast}$ then $V^{\ast}$ is dualizable with dual $V$; moreover, if the ambient symmetric monoidal category is closed, then you can always take $V^{\ast} = [V, 1]$ to be the internal hom of $V$ and the unit, and $\text{ev}$ to be the evaluation map. From here it's not hard to show that

$$V \to [[V, 1], 1] \cong (V^{\ast})^{\ast}$$

is an isomorphism for any dualizable $V$; in other words, dualizable objects are reflexive. It's also true that if $V$ is dualizable then

$$[V, W] \cong V^{\ast} \otimes W$$

for any $W$. None of this requires making any choices and is completely canonical once you show that if duals (together with their evaluation and coevaluation maps) exist then they are unique up to unique isomorphism.

Now, in the case of vector spaces, the dualizable objects are precisely the finite-dimensional ones. Whether you need to exhibit a basis to prove this depends on whether your definition of finite-dimensional is "has a basis." Here it is the coevaluation map which it is troublesome to define without a choice of basis; with a choice of basis $e_i$ it's given by

$$\text{coev} : 1 \ni 1 \mapsto \sum e_i \otimes e_i^{\ast} \in V \otimes V^{\ast}$$

where $e_i^{\ast}$ is the dual basis. But this does not weaken the naturality of anything that's been said here, again because of the important fact that duals (together with their evaluation and coevaluation maps) are unique if they exist.

As in Oscar Cunningham's answer, if your definition of "finite-dimensional" happens to be "the natural map $V \otimes V^{\ast} \to [V, V]$ is an isomorphism" (this map always exists and requires nothing to define) then you can define the coevaluation map to be the inclusion of the identity element into $[V, V]$. At some point you will want to show that something is finite-dimensional, though, and then you might need to pick a basis. But that's okay.