The answer is yes, for I claim that every maximal chain has size continuum.

Suppose that $C$ is a chain of subsets of $\mathbb{N}$ which is
maximal with respect to almost inclusion. Let's work in the
quotient, so we consider only one member from each equivalence
class.

First, notice that $C$ must be dense as a linear order, since
otherwise, it will make a discrete step from some $A$ up to $B$, with
$A\subset^* B$ and $B-A$ infinite, but no member from $C$ is strictly between
$A$ and $B$ modulo finite. In this case, we can make a larger chain by adding
half of $B-A$ to $A$ and creating a new member, violating the maximality of $C$. So it must be dense.

Second, I claim that if $$A_0\subset^*A_1\subset^*\cdots\quad\cdots\subset^*B_1\subset^*B_0$$ are sets
from $C$, with the $A_n$'s increasing from below and the $B_m$'s
decreasing from above, then there must be an element of $C$ above
all the $A_n$'s and below all the $B_m$'s. This is just like
filling an $(\omega,\omega)$ Hausdorff gap. By maximality, $C$
must fill such a gap.

But now, the point is that this property already implies that $C$
must have size continuum. You can find a subset of $C$ ordered
like $\mathbb{Q}$, and then construct such gaps for every real
number, which must be filled by a member of $C$.