If we assume ZF with $\neg$ AC, is the collection of all groups a set? I have learned that the Axiom of Choice is equivalent to the statement that every set can be endowed with a group structure. Now, in searching for answers to the question asked in the title, I have found that the canonical explanation for why the collection of all groups is a proper class is that, for any set (or really for any cardinality) there is a free group on the elements of that set. I'm wondering:
1)Is the fact that we have so many free groups related to/dependent on AC?
2)Are the groups that we lose in switching from AC to $\neg$ AC enough that the collection of groups becomes a set?
 A: No, Choice is not needed to show that the class of groups is not a set. There are several ways to see this.


*

*The class of one-element sets is not a set - and each one-element set can be given a group structure (the trivial group), so there are proper-class-many trivial groups.

*If you want a proper class of nonisomorphic groups, this can still be done: as long as $X$ is an infinite well-orderable set, we can construct (without Choice!) a free group with domain $X$. The claim now follows from Burali-Forti.

*And even this is unnecessary: For $X$ a set, we may - again, without choice - form the free group generated by $X$. The domain of this group will be the set of appropriate equivalence classes of finite strings from $X$, which exists without Choice. And now we just need to find a proper class of sets which yield nonisomorphic free groups in this way, and this is not hard to do. In particular, this is the method you mention in your question - and so the answer to your sub-question is No, Choice is not needed to construct the free group on a set. Note that this group is different from a free group with domain equal to a given set: showing that every infinite set is the domain of a free group does indeed require Choice.
These general principles apply to practically every kind of structure: except in very artificial cases, the class of structures of a certain type is a proper class, and unless there is a bound on the size of the cardinalities of these structures (e.g. finite groups) then there is a proper class of non-isomorphic structures of this type.

Also, your conception of the role of Choice in the size of the universe is incorrect. You can think of a model with Choice as being small, actually, since it doesn't have any non-well-orderable sets; and this makes as much sense as thinking of a model without choice as being small, since it doesn't have enough choice functions. And remember that choice holds in the constructible universe, so any model of ZF contains a submodel of ZFC; so in a precise sense, you can gain choice by losing sets.
A: Consider, in $ZF$, Godel's proper class $L$ of constructible sets. $(L,\epsilon)$ satisfies $ZFC.$
As for $On,$ the ordinals, we have $x\in On\iff (x\in On)^{(L,\epsilon)}.$
For $x\in On$ let $G_x$ be such that  $(G_x$ is a group-structure on $x)^{(L,\epsilon)}.$
Observe that $(G_x$ is a group-structure on $x)^{(L,\epsilon)}\implies (G_x$ is a group structure on $x)$.
So regardless of $AC$ or $\neg AC$, there is a group-structure on every ordinal.  
A: Every singleton can be endowed with a group structure. Even if you want them to be non isomorphic, every power set can be given a [commutative] group structure, using the symmetric difference as the addition operation. 
No choice is needed to show there is a proper class of singletons or a proper class of power sets. 
