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.

totaldomination. Assuming CH, this is also the best you can do for eventual domination because of the Erdos-Rado theorem. I'm not sure what happens if you don't assume CH. – François G. Dorais♦ May 24 '13 at 19:43