Finite sets have a nice "strong self-counting" property: if $X$ is finite, there is a linear order on $X$ such that every $n<\vert X\vert$ is the cardinality of some initial segment of that order. (OK, every order has this property, but meh.)
Now no infinite Dedekind-finite set can be strongly self-counting in this sense. One way to see this is to note that an appropriate linear ordering would have to have initial segments of length $n$ for each $n\in\omega$, and hence an initial segment of ordertype $\omega$. However it's not clear to me that the following weaker property is impossible:
Is there an infinite Dedekind-finite set $X$ and a linear order $\triangleleft$ on $X$ such that every set $Y$ with $\vert Y\vert<\vert X\vert$ is equinumerous${}$-mod-finite to some $\triangleleft$-initial segment of $X$?
("$\vert \cdot\vert$" is used in the choiceless sense.)
Note that if $\vert X\vert$ and $\vert Y\vert$ are equinumerous-mod-finite, we have $\vert X\vert\le\vert Y\vert$ or $\vert Y\vert\le\vert X\vert$: we may WLOG assume $\min\{m,n\}=0$. So a self-counting infinite Dedekind-finite set would give a linearly ordered initial segment of the poset Dedekind-finite cardinalities, which per a hard-to-find paper of Ellentuck could well have strong structural consequences - see the introduction of this paper of Sageev. It seems reasonable at first to hope that in a model where infinite Dedekind-finite cardinalities exist and are linearly ordered by size (which Sageev's paper shows can happen, granting an inaccessible cardinal) every Dedekind-finite set would be self-counting; however, I don't actually see how to prove that.