Obtaining an expression for $\mbox{pos}(n;3)$ is somewhat easier than obtaining the analagous expression for the number of isomorphism classes. In the case at hand, the number of non-identical uniform-height-$3$ posets is obtained by the following reasoning:
Consider the "middle" set of numbers $\{b_m\}$ such that $\forall m :( (\exists a: a < b_m) \wedge (\exists c: b_m < c))$. In a graphical picture, there are three rows of numbers, with $<$ arrows connecting top row numbers to middle row numbers and middle row numbers to bottom row numbers; $\{b_m\}$ is the middle row,
and we can label its count $b$.
Let there be $a$ values of $x: x < b_m$ for some $m$, and $c$ values of $x: x < b_m < x$ for some $m$; so that $n=a+b+c$. Then to form a uniform-height-$3$ poset with row structure $(a,b,c)$ we must follow these rules:
Select which $b$ elements will be in the middle row, and which $a$ elements will be in the top row. This gives an overall factor of $\binom{n}{b}\binom{n-b}{a}$
Each of the $a$ top-row elements must connect (via a $<$ relationship) with some non-empty subset of the middle row elements; this provides a factor of
$(2^b-1)^a$.
None of the $b$ elements may be devoid of any connections to the $a$ row. This subtracts, from the previous factor, $(2^{b-1}-1)^a$. However, this double subtracts cases where two elements of the $b$ row are devoid of these connections. Overall, imposition of the condition that for all $b_m$ there must be an $a$ such that $a<b_m$ turns the previous factor into
$$
\sum_{j=0}^b (-1)^j \binom{b}{j}(2^{b-j}-1)^a
$$
Similarly, the possibilities of connections between the middle and bottom rows provides a factor of
$$
\sum_{k=0}^b (-1)^k \binom{b}{k}(2^{b-k}-1)^c
$$
Correction based on comment by OP
- After specifying the relationships as above, every chain is of length $3$. Since the definition of uniformly high does not preclude some connections of length $2$, each element of row $a$ may have a direct relationship with any subset of the elements of row $c$. This gives a further factor which must be smaller than $2^{ac}$, because the direct connections which replicate indirect connections must not be counted. (That is, there is no difference between $(1<2,2<3)$ and $(1<2,2<3,1<3)$.)
It is difficult to account for the possibility of direct connections (forming chains of length $2$), and I therefore have no results for the problem as posed.
For posets in which every chain is of length precisely, $3$,
combining these factors, and summing over all tuples $(a,b,c)$ that add to $n$, we obtain
$$
\mbox{posExact}(n;3) = \sum_{a=1}^{n-2} \,\,\,\sum_{b=1}^{n-1-a}\left[
\binom{n}{b}\binom{n-b}{a}
\left( \sum_{j=0}^b (-1)^j \binom{b}{j} (2^{b-j}-1)^a \right)
\\\left( \sum_{k=0}^b (-1)^k \binom{b}{k} (2^{b-k}-1)^{n-a-b} \right)
\right]
$$
The next observation is that for $n>37$, $$\log\left(\mbox{posExact}(n;3) \right) < n^2/5$$. Using the usual Stirling approximation tricks, one can show that for sufficiently large $n$,
$$\frac{\log\left(\mbox{posExact}(n;3) \right)}{n^2}$$ is a decreasing function of $n$; in this case, "sufficiently large $n$ is $n>7$ (the peak is reached at $n=7$). Apparently, for all $n\geq 3$, $\log\left(\mbox{posExact}(n;3) \right) > n^2/6$ but I can't seem to prove that. (It is true at least up to $n=800$.)
Numerically, for sufficiently large $n$, it appears that
$$
e^{0.1719n^2} < \mbox{posExact}(n;3) < e^{0.1733n^2}
$$
