Is there a largest large cardinal? In ZFC,  a cardinal is an isomorphism class of sets. However ZFC doesn't explicitly have classes; NBG, which is a conservative extension of ZFC does. 
There is no largest cardinal by Cantors Theorem
There is no set of all sets - it is in fact a class.
Classes do not have cardinalities, as these have been only defined for sets - but if one could define a cardinality for classes - wouldn't this, in some sense be a 'limit' of all cardinals in Set, including all large cardinal axioms?
Thus, is it possible to extend the notion of cardinality in any significant way to classes?
Apologies for the loose phrasing of this question. It was originally going to be a posting on Philosophy.SE, but I thought I would get better answers here.
 A: 
If one could define a cardinality for classes - wouldn't this, in some sense be a 'limit' of all cardinals in Set...

Sure, if you modify the definitions to allow this then the cardinalilty of any proper class would be considered the largest cardinal.  (Cantor's proof would then only apply to cardinals that are sets.)  However, this is not very interesting.

... including all large cardinal axioms?

No, large cardinal axioms are quite different objects from cardinals, and the relationship between largeness of cardinals and "largeness" (strength) of large cardinal axioms is not as straightforward as you might think.  (And anyway, there is no largest large cardinal axiom because there is no maximal consistent recursive extension of $\mathsf{ZFC}$.) 
Many examples of large cardinal axioms take properties of the class of cardinals, e.g. infinitude, regularity (closure under limits of short sequences,) and closure under power sets, and posit that some (set-sized) cardinal already has these property.  Large cardinal axioms therefore provide a sense in which the set/class distinction is made relative, rather than absolute.
This is why I said that the possibility of considering the size of a proper class as a "largest cardinal" is not interesting; if your model of $\mathsf{ZFC}$ is realized as $V_\kappa$ where $\kappa$ is an inaccessible cardinal in my model of $\mathsf{ZFC}$, then a proper class in your model is a set in my model.  Moreover, I can go even farther and define $V_{\kappa+1}$, $V_{\kappa+2}$, etc. So the large cardinal axiom "there is an inaccessible cardinal" transcends the notion of cardinals in $\mathsf{ZFC}$ to a much greater extent than the "one-step" extension of the notion of cardinals that you get by allowing proper classes.
A: Yes, you can extend the notion of cardinality to classes, and it is consistent that there is only one cardinal for classes (e.g. $V=L$ implies that all classes have the same size), or it is consistent that there are classes which are incomparable (e.g. if you add two Cohen subsets to a proper class of cardinals, the class of the pairs cannot be definably well-ordered and it is incomparable in size with the class of ordinals). If you agree to violate the axiom of choices then there are even more options here (classes incomparable with sets).
But I don't think there is a lot written on this topic, and it's scattered throughout many papers. Some obvious consequences here, some minor mentioning there.
As for being a large cardinal, there are two type of properties to consider here. Small properties, which only affect the sets below the cardinal (e.g. inaccessible cardinals) and large properties, which affect sets which are not smaller than the cardinals (e.g. measurable cardinals).  There are also very large properties which affect pretty much everyone in the universe (e.g. supercompact cardinals).
Since $\sf ZFC$ has a very limited access to proper classes, having the class ordinals as a large cardinal of a large property is meaningless. But you can easily have small properties by taking $V_\kappa$ for some $\kappa$ satisfying the wanted property and considering it as a model on its own. If $\kappa$ is a Mahlo cardinal, then in $V_\kappa$ every closed and unbounded set of ordinals includes an inaccessible cardinal.
Note that requiring that every class of ordinals which is closed [and unbounded] has an inaccessible cardinal actually requires less than a Mahlo ordinals and cutting off the universe, because we have less classes to worry about.
Another peculiar example is a Woodin cardinal. Being a Woodin cardinal is having a small property, which itself is a very large property (if $\kappa$ is a Woodin cardinal, it might not even be weakly compact, but $V_\kappa$ is incredibly rich in internally large cardinals). So you can think of the similar case where the ordinals behave like a Woodin cardinal to some extent. Although, I'm not sure why you would do that.
A: Being a large cardinal isn't a question of mere size, is a question of complexity.
Consider $\lambda<\kappa$ with $\lambda$ measurable and $\kappa$ simply strongly inaccesible. Then, $V_\kappa$ is a model of $\mathsf{ZFC}$, $\lambda$ is measurable in this model and as $|V_\kappa|=\kappa$, $\kappa$ is the largest "cardinal" in your sense, but from the outside $\kappa$ isn't very big (isn't measurable).
A: Well, currently, I believe the "largest" large cardinal axiom is "there exists a limit club Berkeley cardinal" i.e. "there exists a regular cardinal $\kappa$  so that $S(\kappa)$ and $\kappa$ is a limit of regular cardinals $\lambda$ so that $S(\lambda)$". Here, $S(\mu)$ is an abbreviation for "$\forall M \ni \mu$ and $C \subseteq \mu$ ($C$ is closed unbounded), there is an elementary embedding $j: M \to M$ with critical point $C$". Of course, one could naively extend beyond this point, but, so far, no "original" large cardinals which have a higher consistency strength than this have been found.
