The Hartogs number of an $n$-element set is $n^{+}$

Transport of well-orders along an injection: let $\alpha$ be an ordinal, let $j:\alpha \to A$ be injective, put $S=j[\alpha ]$ and $R=\{(j(\xi ),j(\eta )):\xi ,\eta \in \alpha ,\xi \in \eta \}$. Then $R$ is a well-order of $S$, the map $j$ is an order isomorphism of $(\alpha ,\in )$ onto $(S,R)$, and $\mathrm{o}\mathrm{t}(S,R)=\alpha$. Indeed $j$ is a bijection onto $S$; $R$ is irreflexive because no ordinal satisfies $\xi \in \xi$; $R$ is transitive as a relation and trichotomous because membership on $\alpha$ is; and a nonempty $T\subseteq S$ has $j^{-1}[T]$ nonempty in $\alpha$, whose $\in$-least element $\xi$ gives an $R$-least element $j(\xi )$ of $T$, since any $j(\eta )\in T$ with $j(\eta )Rj(\xi )$ would have $\eta \in \xi$ and contradict minimality. By construction $j(\xi )Rj(\eta )$ if and only if $\xi \in \eta$, using injectivity of $j$, so $j$ is an order isomorphism and [F8] gives $\mathrm{o}\mathrm{t}(S,R)=\alpha$.

For every ordinal $\alpha$: $\alpha ⪯A$ if and only if $\alpha \le n$. First, $A\approx n$, so an injection $\alpha \to A$ composed with a bijection $A\to n$ gives an injection $\alpha \to n$, and conversely an injection $\alpha \to n$ composed with a bijection $n\to A$ gives an injection $\alpha \to A$; hence $\alpha ⪯A$ if and only if $\alpha ⪯n$. Next $\alpha ⪯n$ if and only if $\alpha \le n$: if $\alpha \in n$ or $\alpha =n$ then $\alpha \subseteq n$ by [F2], so the inclusion is an injection; conversely assume $\alpha ⪯n$ and suppose $\alpha \le ̸n$, so that $n\in \alpha$ by the trichotomy of [F3]. Then $n\subseteq \alpha$ because $\alpha$ is transitive, so $n⪯\alpha$ by inclusion, and [F4] with $\alpha ⪯n$ gives $n\approx \alpha$. If $\omega \le \alpha$ then $\omega \subseteq \alpha$ by [F2], so $\omega ⪯\alpha ⪯n$, while $n⪯\omega$ because $n\in \omega$ and $\omega$ is transitive; [F4] then gives $\omega \approx n$, contradicting [F7]. If instead $\alpha \in \omega$, then $\alpha$ is a natural number by [F6], and [F7] applied to $n\approx \alpha$ gives $\alpha =n$, so $\alpha \le n$, contrary to $n\in \alpha$ and [F2]; hence the supposition fails and $\alpha \le n$.

The set $\{\alpha :\alpha \le n\}$ equals $n^{+}$: $\alpha \le n$ means $\alpha \in n$ or $\alpha =n$ by [F3], that is $\alpha \in n\cup \{n\}=n^{+}$ by [F2], and $n^{+}=n\cup \{n\}=\{0,1,\dots ,n\}$ by the von Neumann convention of [F6].

$\mathrm{\aleph }(A)=\{\alpha :\alpha \text{ is an ordinal and }\alpha ⪯A\}$. One inclusion: if $\alpha =\mathrm{o}\mathrm{t}(S,R)$ for $S\subseteq A$ well ordered by $R$, then the order isomorphism $\alpha \to (S,R)$ of [F8] is injective with image $S\subseteq A$, so $\alpha ⪯A$. The other: if an ordinal $\alpha$ satisfies $\alpha ⪯A$, witnessed by an injection $j$, then step 1.1 produces $S=j[\alpha ]\subseteq A$ and a well-order $R$ of $S$ with $\mathrm{o}\mathrm{t}(S,R)=\alpha$, so $\alpha$ belongs to the set displayed in [F1].

Therefore $\mathrm{\aleph }(A)=n^{+}$: step 2.1 writes $\mathrm{\aleph }(A)=\{\alpha :\alpha \text{ is an ordinal and }\alpha ⪯A\}$ and step 1.2 turns this into $\{\alpha :\alpha \le n\}$, which is $n^{+}$ by step 1.3. Since $\mathrm{\aleph }(A)$ is the least ordinal not injecting into $A$ by [F1], the ordinal $n^{+}$ is that least ordinal, and $\mathrm{o}\mathrm{t}$ of the subsets of $A$ are exactly the ordinals up to $n$.

If $B\approx A$ then $\mathrm{\aleph }(B)=\mathrm{\aleph }(A)$: the hypothesis gives $B\approx n$ as well, so the same computation yields $\mathrm{\aleph }(B)=n^{+}$. Thus the cardinality of a finite set determines its Hartogs number, which is the successor ordinal $n^{+}$ of its finite size $n$; nothing was selected at any stage, since the bijections and order types are given or unique by [F8], so no choice principle is used.