Defining the determinant definitely does not require choosing a basis (though I don't know how you'd define it without at least knowing that a basis exists, for reasons you can see below).

Let's say you've got a linear transformation T:V->V, and suppose V is n-dimensional. Consider exterior powers of V, Λ^k(V); for each one we naturally get a linear transformation Λ^k(T):Λ^k(V)->Λ^k(V). In particular take k=n, so we get a linear transformation Λ^n(T):Λ^n(V)->Λ^n(V). But Λ^n(V) is one-dimensional, so Λ^n(T) must be multiplication by some constant factor. That factor is the determinant.

That's basically the most "natural" definition of the determinant (notice how multiplicativity immediately falls out of it, assuming of course you already know that Λ^k is a functor). You need the idea of a basis in order to (a) make sense of the statement "V is n-dimensional" and (b) prove that Λ^n(V) is 1-dimensional, but that's it, and you certainly never need to choose one in order to define the determinant.

Probably not a great definition for beginning students of linear algebra, but I do have to correct this idea that defining the determinant requires choosing a basis...