Why is there no generalization of the determinant to infinite dimensional vector spaces?

My suspicion is that there are issues with solely considering purely algebraic infinite-dimensional vector spaces and trying to generalize the determinant from its algebraic construction.

However, there are analytic generalizations of determinants. Some of them are very deep and I am no expert on them, but there is at least one very concrete generalization I am familiar with.

Definition: Let $A$ be a finite rank operator on a Hilbert space. Then define $$\det(I - A) = \prod_j (1 - \lambda_j(A)),$$ where $\lambda_j$ is the $j$th largest eigenvalue of $A$. If $A$ is instead a trace class operator on a Hilbert space, then define $$\det(I - A) = \lim_{k \rightarrow \infty} \det(I - A_k),$$ where $\{A_k\}$ is a sequence of finite rank operators such that $\|A - A_k\|_{L^1} \rightarrow 0$.

Since you already seem satisfied with the argument that the sign homomorphism does not extend to the group of (not necessarily finitely supported) permutations of $\Bbb N$, consider that by acting on the canonical basis of $K[X]$, such permutations define an isomorphic subgroup of automorphisms of this infinite dimensional vector space, and that an extension of the determinant to $\operatorname{Aut}_K(K[X])$ would by restriction to that subgroup define such an impossible extension of the sign homomorphism to permutations of $\Bbb N$.

The answer to your question depends exactly on what you want the determinant to have.

A reasonable set of properties to ask on a determinant is that $$ \tag{0} \det\mathrm{id}_V = 1 $$ for all identity maps $\mathrm{id}_V:V\to V$ of all vector spaces $V$, that $$ \tag{1} \det(f\circ g)=\det f\cdot\deg g $$ whenever $f:V\to V$ and $g:V\to V$ are linear maps, and that $$ \tag{2} \det\begin{pmatrix}f&h\\0&g\end{pmatrix}=\det f\cdot\det g $$ whenever $f:V\to V$, $g:W\to W$ and $g:W\to V$ are linear maps (the matrix on the left denotes the hopefully obvious linear map $V\oplus W\to V\oplus W$).

Suppose we do have such a map and let $f:V\to V$ be an automorphism of a vector space $V$. Let $V^\infty=V\oplus V\oplus V\oplus\cdots$ be a countable sum of copies of $V$ and $f^\infty=f\oplus f\oplus f\oplus\cdots$ the obvious automorphism of $V^\infty$. Using only the above two properties of the function~$\det$ you can check that $\det f^\infty$ is a non-zero element of the field and that $$\det f\cdot\det f^\infty=\det f^\infty,$$ so that in fact $$\det f=1.$$

The conclusion of this is that if you want the define the determinant of automorphisms of vector spaces in such a way that (0), (1) and (2) hold, then you only can define it to be always $1$, not the most interesting definition.