Let $B\_h$ be the *perfect* binary tree having height $h$ (i.e. the binary tree with height $h$ in which all interior nodes have two children and all leaves have the same depth or same level). 


For any rooted tree $T$, we denote by $r(T)$ its root. For instance, $r(B\_3)$ is the root of the perfect binary tree having height $3$.


Let $F\_{h,m}$ be the set of *all* possible forests that ***(i)*** are formed by at most $m>1$ trees and ***(ii)*** can be obtained by removing from *any* tree $B\_{h}$ (for all $h>1$) all nodes (together with their incident edges in $B\_h$) of any subtree $T'$ of $B\_h$ rooted at $r(B\_h)$ (thus we have $r(T')=r(B\_h)$).




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**Question**: How can we calculate (or bounding from above) the cardinality of $F\_{h,m}$ asymptotically when $h\to\infty$ (as a function of $m$ and $h$)? Does the bound $|F\_{h,m}|\le\sum\_{i=0}^m \binom{2^h}{i}$ hold for all $h, m>1$?



Counting the forests obtainable by removing subtrees from binary trees
It is convenient to consider the vertex set of $B\_h$ as a partially ordered set with its natural genealogic order, with minimum element its root. Any subtree $T'$ as described in your procedure (also including the empty tree) is then exactly an *initial segment*. (For the notation: $B\_1$ is the one-vertex tree, and so on)


Removing $T'$ from $B\_h$ leaves a set of $m$ complete trees, whose $m$ roots constitue exactly an $m$-*antichain* in the partial order. The only $0$-antichain is the empty set and produces the empty forest. The $1$-antichains are the vertices as singletons; there are $2^{h}-1$ of them, including the root. The number of forests with *exactly* $m$ trees in $B\_h$ obtained removing a $T'$ is therefore the number $a\_{h,m}$ of all $m$-antichains in $B\_h$. So in your question $|F\_{h,m}|=\sum\_{2\le j\le m}a\_{h,j}$, since you like to consider forests with more than one, and no more than $m$ trees. There is a clear convolution formation passing from antichains in $B\_h$ to antichains in $B\_{h+1}$, giving the recursive relation for the generating polynomial $P\_h(x):=\sum\_{m\ge0}a\_{h,m}x^m$, $$\cases{P\_{0}(x)=1\\P\_{h+1}(x)=x+P\_h(x)^2.}$$


(and of course $P\_{h}(x)/(1-x)$ counts antichains with at most $m$ elements in $B\_h$). The problem of counting antichains in posets is well-studied. Here is a 2006 paper "[*Counting chains and antichains in the complete binary tree*](https://www.researchgate.net/publication/220620320_Counting_chains_and_antichains_in_the_complete_binary_tree)" 
you may want to request to the authors (Though I'm not sure if it address the $m$-element version). Incidentally, note that $P\_h(1)$, the number of all antichains in $B\_h$, is the sequence [A003095](https://oeis.org/A003095).


**Edit.** In fact, if you want an asymptotics for $a\_{h,m}$ as $h\to+\infty$ for each fixed $m$, this is easily available: It follows by induction that for any $m$
$$a\_{h,m} \sim {2^{mh}\over m!} \quad\text{ as }\;h\to+\infty, $$
so that the bound given by Emil Jeřábek in comment below in terms of $m$-elements subsets of $B\_h$ is also an asymptotic, since
$$ {2^{mh}\over m!} \sim { 2^{h}-1\choose m } \quad\text{as }h\to+\infty. $$ Indeed we have, for $m=1$:
$$a\_{h,1} =2^{h}-1\sim 2^{h}\quad\text{as }h\to+\infty,$$
while for $m>1$,
$$a\_{h+1,m}=\sum\_{j=0}^m a\_{h,j}a\_{h,m-j}= 2a\_{h,m}+\sum\_{j=1}^{m-1} a\_{h,j}a\_{h,m-j}$$
so assuming by complete induction hypothesis that $a\_{h,j}\sim {2^{jh}\over j!}$ as $h\to+\infty$ holds for each $ j<m$, we find
$$a\_{h+1,m}- 2a\_{h,m}\sim {2^{mh}\over m!}\, \sum\_{j=1}^{m-1} {m!\over j!(m-j)!} \,=\,{ 2^m-2\over m!} \,2^{mh}$$
or, dividing by $2^ {h+1 }$
$${a\_{h+1,m}\over2^{ h+1 }}- {a\_{h ,m}\over2^{ h }}\sim { 2^{m-1}-1\over m!}\,2^{(m-1)h} $$
whence, summing from $0$ to $h-1$
$${a\_{h ,m}\over 2^{ h }}\sim {2^{m-1}-1\over m!}\,\sum\_{j=0}^{h-1}2^{(m-1)j}={2^{(m-1)h}-1\over m!}\sim{2^{(m-1)h}\over m!} $$
so that 
$$a\_{h,m} \sim {2^{mh}\over m!} \quad\text{as }h\to+\infty.$$


