What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$? Let $\mathbb{R}^\mathbb{R}$ be the set of functions $\mathbb{R}\to\mathbb{R}$ patially ordered by eventual domination. Obviously, every ordinal below $\omega_1$ can be embedded in $\mathbb{R}^\mathbb{R}$ using only constant functions.
What is the least ordinal than cannot be embedded in $\mathbb{R}^\mathbb{R}$?
 A: One can define an eventually increasing sequence $\{f_\alpha:\alpha<\omega_1\}$ by transfinite recursion on $\alpha$. By inserting sequences in the intervals $(f_\alpha(x),f_{\alpha+1}(x))$ we can see that $\omega_1$-sums (and of course, $\omega$-sums) do not lead out of the representable ordinals. This gives that all ordinals $<\omega_2$ are representable and this is clearly sharp if CH holds, by Will's above answer. 
If CH fails, larger ordinals can be represented, if, e.g. Martin's Axiom holds. 
A: Let me get things started with some simple observations.
Note that given any countable sequence of functions $f_n$, we can
by diagonalization construct a function eventually dominating all
of them, $f(x)=\max_{n\leq x}f_n(x)$. It follows that we may by
transfinite recursion construct an embedding of $\omega_1$ into
your order: at successor stages, add one to the previous function;
at limit stages, use the diagonalization just described.
So actually, since $\mathbb{R}$ is order-isomorphic to bounded
intervals of itself, we can therefore also embed $\omega_1$ into
the order many times, on top of one another. So this gives
strictly larger ordinals mapping in.
More generally, the bounding number $\mathfrak{b}$ is the size
of the smallest unbounded family of functions, and any family of
size less than $\mathfrak{b}$ will be bounded above. Thus, the
recursive construction actually shows that we can find an
embedding of $\mathfrak{b}$ into $\mathbb{N}^{\mathbb{N}}$ under
eventual domination. Thus, we also get strictly larger ordinals
than $\mathfrak{b}$ embedding in, by using the bounded-interval trick again.
There are diverse independence results concerning the exact
value of $\mathfrak{b}$. Under CH, it is the same as the
continuum, of course, but when CH fails, it can be far larger than
$\omega_1$.
Using Péter's idea, once we have a map from $\mathfrak{b}$ into the order, then we may conclude that the class of ordinals that map into the order is closed under sums of length $\mathfrak{b}$. Thus, any ordinal up to $\mathfrak{b}^+$ is is order-embeddable into $\mathbb{R}^\mathbb{R}$ under eventual domination. So $\mathfrak{b}^+$ is a lower bound for your desired ordinal.
I guess the same idea shows that whenever an ordinal $\kappa$ embeds in, then the class of ordinals will be closed under sums of length $\kappa$, and so all ordinals up to $\kappa^+$ will also map in. Thus, the smallest ordinal not embedding in must be a cardinal, and furthermore, it must be a regular cardinal for the same reason.
Update.  It is relatively consistent that the answer is $\mathfrak{c}^+$, even when the continuum $\mathfrak{c}$ is very large, and much larger than $\mathfrak{b}$. The reason is that by forcing, we can undertake a very long forcing iteration of length $\kappa$ to add a dominating real at each stage, and thereby get a model with continuum $\kappa$, such that $\kappa$ embeds into the order (and so the smallest ordinal not embedding into the order is $\kappa^+$). Now, the point is that with further ccc forcing, we can make $\mathfrak{b}$ small or whatever we like, but meanwhile, we still have our old functions showing that $\kappa$ maps into the order. 
