I claim that this topology has the stronger property of being hyperconnected, i.e., the intersection of any two nonempty open sets is nonempty.
Indeed, any open set contains a set of the form
$$ U=\bigcap_{i=1}^m \{x:x\not\le^*y_i\}\bigcap\bigcap_{j=1}^n \{x:x\not\ge^*z_j\} $$
so the same can be said of the intersection of two open sets. I claim that
$U$ is nonempty iff every $z_j$ has infinitely many nonzero terms.
It is easy to see this claim implies that the topology is hyperconnected. Now we show the claim.
Clearly this condition is necessary, for if $z_j$ has finitely many nonzero terms, then for every $x\in\mathbb N^{\mathbb N}$ we have $x\ge^*z_j$. Conversely, assume all of $z_j$ has infinitely many nonzero terms, then we can produce an $x\in U$ by the following algorithm:
x=0 //initialize x to the zero array
k=0 //k indexes x
while true do
for i = 1 to m
x(k)=y(i)(k)+1 //Repeating this infinitely makes x\not\le^*y_i
k=k+1
end for
for j = 1 to n
while z(j)(k)=0 do
k=k+1
end while //This loop must end because z(j) has infinitely many nonzero terms
x(k)=0 //Repeating this infinitely makes x\not\ge^*z_j as z(j)(k)>0
end for
end while
Edit: As Todd noted in the comment, another natural definition of the interval topology is generated by the open sets $\{x:x^*>y\}$ and $\{x:x^*<z\}$, where $x<^*y$ iff there is $n_0$ such that for all $n>n_0$, $x_n<y_n$.
This topology is very different from the previous one. In fact it is not connected. It can be written as $U\cup V$, where $U$ is the set of all sequences tending to $+\infty$ and $V$ is its complement. We will show that both $U$ and $V$ are open.
Take $x\in U$. Then there is $n_0$ such that for all $n>n_0$, $x_n>0$. Let
$$ y_n=\begin{cases}x_n & n\le n_0\\ x_n-1 & n>n_0 \end{cases}. $$
Then $y\in U$ and $x\in (y,+\infty)\subset U$.
Take $x\in V$. Let $z=x+1$ (termwise). Then $z\in V$ and $x\in (-\infty,z)\subset V$.
