Order dimension of $\omega^\omega/(fin)$ Let $\omega^\omega$ denote the collection of all functions $f:\omega\to\omega$. For $f,g\in\omega$ we say $f\simeq g$ if and only if $\exists N \in \omega$ such that $f(n) = g(n)$ for all $n\geq N$. 
Often, the symbol $(fin)$ is used for the equivalence relation $\simeq$. For $[f], [g] \in \omega^\omega/(fin)$ we define $[f]\leq^* [g]$ if and only if $\exists N\in\omega$ such that $f(n)\leq g(n)$ for all $n\in \omega$ with $n\geq N$. Clearly, $\leq^*$ is well-defined and $(\omega^\omega/(fin), \leq^*)$ is a poset.
What is the order dimension of $(\omega^\omega/(fin), \leq^*)$?
 A: $\newcommand\Fin{\text{Fin}}$Theorem. The order dimension of
$\langle\omega^\omega/\Fin,\leq^*\rangle$ is precisely the
continuum.
Proof. It is easy to see that the dimension is at most the
continuum, since the space itself has size continuum. (One must
argue that for any instance of incomparability, we may find two
linear orders extending $\leq^*$ placing the two individuals in
opposite order, so that the intersection witnesses that instance
of incomparability.)
Let me now argue that the dimension is at least continuum. For
this, I build a continuum-sized version of the example of
dimension four described on the wikipedia page to which you link. Probably it would help to be familiar with that finite example to understand how the one I present here works, since the underlying idea is the same. 
Specifically, we first construct inside
$\langle\omega^\omega/\Fin,\leq^*\rangle$ functions $a_s,b_s\in\omega^\omega$ indexed by 
$s\in 2^\omega$, such that $s\neq t\to a_s\perp a_t,\ b_s\perp
b_t$ and also $a_s\perp b_s$ and $a_s\leq^*b_t\iff s\neq t$. So these functions are
two antichains, with $a_s$ below all the $b_t$ except for $b_s$.
To construct this situation, let $A_s\subset\mathbb{N}$ be an
almost disjoint family for $s\in 2^\omega$. Such a family can be
constructed by labeling the nodes of the tree $2^{<\omega}$ with
distinct natural numbers, and letting $A_s$ be the labels on the
branch $s$. It follows that each $A_s$ is infinite and $s\neq t\to
A_s\cap A_t$ is finite. Now, let $a_s$ be the characteristic
function of $A_s$, that is, $a_s(n)=1$ if $n\in A_s$ and otherwise
$0$. These functions are pairwise incomparable. Let $b_s$ be the
characteristic function of the complement of $A_s$. These
functions are also pairwise incomparable. Because the family is
almost disjoint, it follows that $a_s\leq^* b_t$ whenever $s\neq
t$, but $a_s\perp b_s$. So the family is as desired.
Now suppose that we have a family of linear orders whose
intersection is $\leq^*$. For any particular $s\in 2^\omega$, then
since $a_s$ and $b_s$ are incomparable by $\leq^*$, there must be
a linear order $<$ in the family for which $b_s<a_s$. But now,
since $a_s\leq^*b_t$ and $a_t\leq^* b_s$ for all $t$, it follows
that this order $<$ works only for this particular $s\in
2^\omega$.
So the family of linear orders must have size at least continuum.
QED
