The question is "really" about quasi-compactness, which is usually assumed as a hypothesis in versions of Zariski's Main Theorem. However, the other strong hypotheses of the OP imply quasi-compactness in this case. The key point is that an open immersion is quasi-compact if the target is Noetherian.
Lemma. Let $i:X\to Z$ be a separated morphism between irreducible schemes. If there exists a covering of $X$ by open affines $U$ such that each restriction $i|_U$ is an open immersion, then $i$ is an open immersion. If $Z$ is Noetherian, then $X$ is quasi-compact.
Proof. Up to replacing $Z$ by the open image of $i$, assume that $i$ is surjective. The goal is to prove that $i$ is an isomorphism. We construct the inverse isomorphism $i^{-1}:Z\to X$ by gluing. Let $U$ and $V$ be nonempty open subschemes of $X$. The cocycle condition for $i^{-1}$ is precisely the condition that $i^{-1}(i(U)\cap i(V))$ equals $U\cap V$.
Let $Y^o$ be a nonempty open affine subset of the open intersection $i(U)\cap i(V)$. Denote by $X^o$ the inverse image $i^{-1}(Y^o)$. Since $X$ is irreducible, the intersections of nonempty open subsets $U\cap X^o$ and $V^\cap X^o$ are dense. Denote these by $U^o$ and $V^o$. By construction, each of the following restrictions of $i$ is an isomorphism, $$i_U:U^o\to Y^o, \ \ i_V:V^o\to Y^o.$$ These isomorphisms agree on $U^o\cap V^o = (U\cap V)\cap X^o$.
Since $i$ is separated and since $Y^o$ is affine, the scheme $X^o$ is separated. Define $j$ to be the automorphism of $X^o$ whose restriction to $U^o$ equals $i_V^{-1}\circ i_U$ and whose restriction to $V^o$ equals $i_U^{-1}\circ i_V$. These glue since $i_U$ and $i_V$ agree on $U^o\circ V^o$. Moreover, $j$ equals the identity on $U^o\circ V^o$. Since $j$ and the identity agree on the dense open $U^o\circ V^o$, and since $X^o$ is separated, the morphism $j$ equals the identity. Thus, $U^o$ equals $V^o$. Since we can cover $i(U)\cap i(V)$ by such open affines, it follows that $i^{-1}(i(U)\cap i(V))$ equals $U\cap V$.
Finally, if $Y$ is Noetherian, then every open subset of $Y$ is quasi-compact. Thus, the scheme $X$ is quasi-compact. QED
Let $f:X\to Y$ be a locally finite type, separated morphism with finite fibers that is quasi-finite Zariski locally on $X$, and that is strongly dominant in the sense that the $f$-inverse image of every dense open subset of $Y$ is a dense open subset of $X$. Assume also that $X$ is normal and that $Y$ is quasi-compact, separated, excellent, integral, and normal.
Proposition.(Variant of Grothendieck's "Zariski Main Theorem") There exists a factorization of $f$ as the composition of a dense open immersion into a normal scheme, $i:X\hookrightarrow Z$, followed by a finite, strongly dominant morphism, $g:Z\to Y$. Moreover, this factorization is unique up to unique isomorphism.
Proof. Every irreducible component of $X$ dominates $Y$, i.e., every generic point of $X$ maps to the generic point of $Y$. By hypothesis, there are only finitely many preimages of the generic point of $Y$, i.e., $X$ has only finitely many irreducible components. Since $X$ is normal, these irreducible components are connected components. Without loss of generality, assume that $X$ is connected, i.e., $X$ has a unique generic point $\eta$.
Since $Y$ is excellent, the "integral closure" of $Y$ in the "function field" $\kappa(\eta)$ is a finite, strongly dominant morphism whose domain is normal, $$g:Z\to Y.$$ By the universal property of the normalization, there exists a unique morphism of schemes compatible with the specified morphisms to $Y$, $$i:X\to Z.$$ By Zariski's Main Theorem, working locally on $X$ with opens that are quasi-compact over $Y$, the morphism $i$ is an locally an open immersion.
By the lemma, the morphism $i$ is an open immersion.QED
Now you can apply this when $Y$ is affine space. I recommend that you read Grothendieck's formulation of Zariski's Main Theorem in EGA.
