A similar result seems to hold also for all higher level nested binomial coefficients eg for level ${{{n \choose t} \choose d_1} \choose d_2}= \sum_{k=1}^{td_1d_2}A_{t,d_1,d_2,k} {n \choose k}$ three,
${{{n \choose t} \choose d_1} \choose d_2}= \sum_{k=1}^{td_1d_2}A_{t,d_1,d_2,k} {n \choose k}.$
Transforming to binomial base It is known (OEIS A19538) that $x^n=\sum_{k=1}^n T(k,n){n \choose k}$ where $T(k,n)=k! S(n,k)$, and $S(n,k)$ is Stirling numbers of the second kind so that we have explicit formula for conversion to binomial base :
The log-concavity of $A_{t,d,k}$ probably just follows from those of $T(k,n)$.
Geometric meaning of the coefficients We are led to consider the nested binomials from a weight formula for enumerating subgraphs of $K_n$. Let $K_n$ be the complete graph on $n$ vertices, a subgraph $H$ is just a subset of the edges (so this ignore and excludes isolated points). We recently found a mass or weight formula for enumerating all such subgraphs. Each subgraph has an induced number of vertices $k=k(H)$ (eg. $k=3$ for triangle and $k=6$ for three disjoint edges). The weight formula for subgraphs of $K_n$ is
where $Aut_k(H)$ is the automorphism group of $H$ as a subgroups of $S_k$ of permutation of the induced vertices. This has a refinement if sum over subgraphs with a fixed number $d$ of edges. (Is this formulasomething well known ?)
If we now let $a_{2,d,k}$ be the sum over all $H<K_n$ on $d$ edges with the number of induced vertices $k(H)=k$ of the weight $w(H)=\frac{k!}{|Aut_s(H)|}$weights,
$ a_{2,d,k}:=\sum_{H<K_n,|H|=d,k(H)=k} \frac{k!}{|Aut_s(H)|},$$(1) \;\;\; a_{2,d,k}:=\sum_{H<K_n,|H|=d,k(H)=k} \frac{k!}{|Aut_s(H)|},$
$(*) \;\; \sum_{k=1}^{2d} a_{2,d,k}{n \choose k}={{ n \choose 2} \choose d}.$$(2) \;\; \sum_{k=1}^{2d} a_{2,d,k}{n \choose k}={{ n \choose 2} \choose d}.$
The upper limit $2d$ is to allow the maximal possible number of induced number of vertices on a union of $d$ disjoint edges but it still holds if $n<2d$ if we follow the usual convention that ${n \choose k}=0$ if $k<n$.
SoThe weight formula (*2) give a constraint on our geometric data $a_{2,d,k}$ but we then realize the constraint determine the data uniquely viewing ${{n \choose 2 } \choose d}$ as a polynomial in $n$ and expressing it in its unique binomial base. We must now have
whichso that can now berewrite (1) as
$\sum_{H<K_n,|H|=d,k(H)=k} \frac{k!}{|Aut_s(H)|}=A_{t,d,k},$
and viewed it as a refined weight formula since the RHS can be defined independently. Every thing still holds if we replace $2$ by $t$