The ordinals are well ordered regardless to any assumption of choice. It is defined that the class of ordinal numbers is the smallest transitive class that can be well ordered, and they form a backbone of transitive models, that is if $M,N$ are two transitive models have the same ordinals then $L^M=L^N$.
Without the axiom of choice there are non-well orderable sets. Their cardinality, if so, is not an $\aleph$ number. We define the cardinality of $X$ as either finite, or $\aleph_\alpha$ for some ordinal $\alpha$, in case that $X$ can be well ordered; or as a definable subset of the class of $A$'s such that there is a bijection between $X$ and $A$.
For example, it is consistent with ZF that there are infinite sets that cannot be split into two infinite sets (every partition into two disjoint sets will yield one of them finite). Such set does not even have a countable subset and therefore incomparable with $\aleph_0$.
To see how total order of cardinals is equivalent to the axiom of choice:
If the axiom of choice holds, then every set can be well ordered and is finite or equivalent to some $\aleph$-cardinal. Therefore all cardinals are $\aleph$'s and so cardinals are totally ordered (and well ordered too).
On the other hand, if cardinals are totally ordered, given a set $X$ denote $H(X)$ the least ordinal $\alpha$ such that there is no injective function from $\alpha$ into $X$.
It can be shown that $\alpha$ is an $\aleph$ cardinal (by the fact it has no bijection with smaller ordinals which are injectible into $X$), and $\alpha\nleq|X|$.
By the assumption that cardinalities of all sets are comparable $|X|$ is comparable with $\alpha$, therefore $|X|<\alpha$, and we have that $X$ can be injected into $\alpha$ and so it inherits a well order from such injection.
Therefore all sets can be well ordered, which is equivalent to the axiom of choice.
(The term for which I used for $H(X)$ is also known as Hartogs number)