Here at least is the usual justification for moving from AC for sets to what is normally called the global axiom of choice, which asserts that there is a class well-ordering of the (first-order) universe.
Theorem.
The global axiom of choice, when added to the ZFC or
GB+AC axioms of set theory, leads to no new theorems about sets.
That is, the first-order assertions about sets that are provable
in GBC are precisely the same as the theorems of ZFC.
Furthermore, every model of ZFC can be extended (by forcing) to a
model of GBC, in which the global axiom of choice is true, while
adding no new sets (only classes).
In particular, the global axiom of choice is safe in the
sense that it will not cause inconsistency, unless the underlying
system without the global axiom of choice was already inconsistent.
Proof. Suppose that $M$ is any model of ZFC. Consider the class
partial order $\mathbb{P}$ consisting of all well-orderings in $M$
of any set in $M$, ordered by end-extension. As a forcing notion,
this partial order is $\kappa$-closed for every $\kappa$ in $M$,
since the union of a chain of (end-extending) well-orderings is
still a well-order. If $G\subset\mathbb{P}$ is $M$-generic for
this partial order, then $G$ is, in effect, a well-ordering of all
the sets in $M$. Furthermore, one can prove by the usual forcing
technology that the structure $\langle M,{\in},G\rangle$ satisfies
$\text{ZFC}(G)$, that is, where the predicate $G$ is allowed to
appear in the replacement and other axiom schemes.
Essentially, what we've done is add a global well-ordering of the
universe generically. And since the forcing was closed, no new
sets were added, and so $M[G]$ has the same first-order part as
$M$.
It follows now that GBC is conservative over ZFC for first-order
assertions, since any first-order statement $\sigma$ that is true
in all GBC models will be true in $M[G]$ and therefore also in
$M$, and so $\sigma$ is true in all ZFC models as well. QED
I think the main reason replacement is seen as an essential part of ZF is that it naturally follows from the ontology of set theory, as do the other axioms of ZF. The ontology of set theory is rooted in the idea that sets are obtained by an iterative process along a wellordered "ordinal clock", where at each step all the sets whose elements were generated earlier now appear. It is intuitively clear that, in order to be exhaustive, this process must go on for a long, long time. From this point of view, the replacement axiom can be intuitively stated: no set can be used as an indexing of a family of ordinals that reaches to the end of the ordinal timeline. This is a natural consequence of the idea that the iterative process should be exhaustive.
Another interesting aspect of replacement is that (in most formulations of ZFC) it is logically equivalent to reflection. (Informally: for any formula $\sigma(\vec{x})$ in the language of set theory, if $V \models \sigma(\vec{x})$ then there is an ordinal $\alpha$ such that $V_\alpha \models \sigma(\vec{x})$.) This is an extremely useful principle. One of its side effects is that ZFC is "self-justifying" in the sense that any finite fragment of ZFC is realized in a level $V_\alpha$ of the cumulative hierarchy. In other words, if one were to test set theory by examining a finite fragment of the axioms within the universe of sets, one would see that this finite fragment is not only consistent but that it has a model $V_\alpha$ that arises from the same iterative process that all sets do. In particular, ZFC comes very close to proving its own consistency even though we know this is not possible after Gödel. This feature makes ZFC very appealing as a foundational theory. (Note that PA has a similar self-justifying feature, but ETCS doesn't appear to have this.)
Another, more practical, use of replacement is to obtain "cheap universes". Grothendieck universes have proven useful for handling large objects. Unfortunately, one cannot prove their existence in ZFC. It is nevertheless often true that a "reasonable theorem" proven using Grothendieck universes is actually provable in ZFC. The reason is that the proof often doesn't make full use of all the features of Grothendieck universe, a finite fragment of those features often suffices and in such cases reflection provides a set $V_\alpha$ with all the necessary features to make the argument work. ETCS doesn't appear to have a good way of obtaining "cheap universes". This also hints at an alternative to replacement, which is to have a hierarchy of universes similar to those used in dependent type theory.
Operations on dependent families is where the need for replacement arises most. It's actually really hard to even talk about dependent families in the language of ETCS. The main issue with ETCS isn't necessarily that it can't prove the existence of coproducts like $\coprod_{n \in \mathbb{N}} \mathcal{P}^n(\mathbb{N})$, but that it has a hard time even talking about the family of all $\mathcal{P}^n(\mathbb{N})$ in the first place. Introducing universes would be an interesting way to get around that problem but there are other means, all of which are likely to make the need for replacement-like principles clear.
As for the proposed workaround, it's unclear you would get much more by this kind of process. Rather than ETCS, I'll work in BZC extended with terms for powersets and union and a constant symbol for $\omega$. The exponential-bounded formulas are defined like bounded formulas except that the bounding terms can involve powerset and union.
Fact. If $\phi(x,y)$ is an exponentially-bounded formula such that BZC proves that $\forall x \exists y \phi(x,y)$ then there is a standard number $n$ such that BZC proves that $\forall x \exists y(\phi(x,y) \land |y| \leq |\mathcal{P}^n(x \cup \omega)|)$.
Proof. Find a model $M$ of BZC and consider $x \in M$. Let $M' = \bigcup_n \{z \in M : |z| \leq |\mathcal{P}^n(x \cup \omega)|\}$ where $n$ ranges over the standard numbers only. Note that $M'$ is model of BZC that contains $x$. By hypothesis, there is a $y \in M'$ such that $M' \models \phi(x,y)$. Note that exponential-bounded formulas are absolute between $M'$ and $M$ since the only sets that we need to look at to figure out that $\phi(x,y)$ is true are in $M'$. Thus $M \models \phi(x,y)$ and also $M \models |y| \leq |\mathcal{P}^n(x \cup \omega)|$ by definition of $M'$. Now the fact that there is a fixed standard $n$ that provably works for all $x$ follows by a compactness argument. $\square$
The examples $\omega, \mathcal{P}(\omega), \mathcal{P}^2(\omega), \ldots$ and $V, V^{\ast}, V^{\ast\ast},\ldots$ are definable by a formula of the form $\exists z\phi(x,y,z)$ where $\phi(x,y,z)$ is exponential-bounded. Because of the nice biinterpretation between ETCS and BZC, for these and similar examples, either ETCS doesn't prove that the $n$-th iterate exists for every natural number $n$, or ETCS already proves that replacement holds in this particular instance.
Let me also address one aspect of the footnote, which states that replacement would be "the most likely culprit" if ZFC were found to be inconsistent. The "standard objection" to axiomatic set theory is actually with comprehension. If you think about it, comprehension is a rather bold statement: every formula in the language of set theory can be used to define a subset of a given set. The issue is that formulas can be complex beyond (human) understanding, it's hard to justify the use of comprehension for formulas we can't understand. In fact, it's not clear that comprehension is fully justified by the ontology of set theory described above. (Note that the same kind objection applies to PA, where one asks for induction to hold for arbitrary formulas.)
Best Answer
There’s one issue underlying a lot of the discrepancies between people’s answers, I think:
(hence also, how we define subsequent things like “a small-category-indexed diagram of sets”) There are at least two main options here:
$f$ is a class of pairs, such that…
$f$ is a set of pairs, such that…
At least in most traditional presentations, I think it’s defined as the latter, but some people here also seem to be using the former. The answer to this question depends on which we take.
If we take the “a function is just a class” definition, then as suggested in the original question, and as stated in François’ answer, we definitely have some big problems without replacement: Set is no longer complete and co-complete, etc. (nor are the various important categories we construct from it); we can’t easily form categories of presheaves; and so on. Under this approach, we certainly get crippling problems in the absence of replacement.
On the other hand, if we take the “a function must be a set” definition, we get some different problems (as pointed out in Carl Mummert’s comments), but it’s not so clear whether they’re big problems or not. We now can form limits of set-indexed families of sets; presheaf categories work how we’d hope; and so on. The problem now is that we can’t form all the set-indexed families we might expect: for instance, we if we’ve got some construction $F$ acting on a class (precisely: if $F$ is a function-class), we can’t generally form the set-indexed family $\langle F^n(X)\ |\ n \in \mathbb{N} \rangle$.
This is why we still can’t form something like $\bigcup_n \mathcal{P}^n(X)$, or $\aleph_\omega$. On the other hand, such examples don’t seem to come up (much, or at all?) outside set theory and logics themselves! Most mathematical constructions that do seem to be of this form — e.g. free monoids $F(X) = \sum_n X^n$, and so on — can in fact be done without replacement, one way or another.
Now… I seem to remember having been shown an example that was definitely “core maths” where replacement was needed; but I can’t now remember it! So if we take this approach, then we certainly still lose something; but now it’s less clear quite how much we really needed what we lost.
(This approach is very close to the question “What maths can be developed over an arbitrary elementary topos?”.)