In the definition of $n$-equivalence, what is the motivation for only requiring surjectivity on the $n$th dimension. An $n$ equivalence $f\colon X \to Y$ such that the induced map on the homotopy group $f_* \colon\pi_m(X) \to \pi_m(Y)$ is an isomorphism for $m<n$ and an epimorphism for $m=n$.
What's the motivation for the weaker assumption in the dimension $n$?  Why do we want to consider it an $n$-equivalence even though the map doesn't give an isomorphism in dimension $n$?
 A: I would call such a map $(n-1)$-connected, but I suppose that is just a matter of terminology. (Incidentally, Lurie calls these $n$-connective.)
Let $n \ge 0$. Recall that an $n$-connected space is a non-empty $X$ such that $\pi_m (X, x) = 1$ for all $x$ in $X$ and all $m \le n$. (We also define a $(-1)$-connected space to be a non-empty space, and a $(-2)$-connected space is any space at all.) An $n$-connected map is, roughly speaking, a map whose homotopy fibres are $n$-connected.
To simplify matters, let us consider a fibration $p : E \to B$. Then the homotopy fibres of $p$ can be computed as the ordinary fibres of $p$, and we have a long exact sequence of homotopy groups,
$$\cdots \to \pi_{m+1} (B, b) \to \pi_m (F, e) \to \pi_m (E, e) \to \pi_m (B, b) \to \pi_{m-1} (F, e) \to \cdots$$
where $b \in B$, $F = p^{-1} \{ b \}$, and $e \in F \subseteq E$. Thus, if $F$ is $n$-connected, then $\pi_m (F, e) = 1$ for all $m \le n$, and by exactness, we have $\pi_m (E, e) \to \pi_m (B, b)$ surjective for $m = n + 1$ and bijective for $1 \le m \le n$; and if every fibre is $n$-connected, then $\pi_0 E \to \pi_0 B$ is also bijective. Conversely, if we know $p : E \to B$ is an $n$-connected fibration, then by considering the long exact sequence we can deduce that the fibres are $n$-connected. Thus, a fibration is an $n$-connected map if and only if its fibres are $n$-connected.
As for "equivalence", I suppose that the point is that an $\infty$-connected map is precisely a weak homotopy equivalence.
