Suppose that the root is on level $n$. If $n=0$ or $n=1$, none of the nodes pays or receives anything, so each node has a balance of $0$. (To avoid clutter I omit the dollar sign.) If $n=2$, the $4$ nodes on level $0$ pay $7$ each and receive nothing, so they end up with a balance of $-7$; the $2$ nodes on level $1$ neither pay nor receive and have a balance of $0$; and the root node receives $4\cdot7=28$ and pays nothing, for a balance of $28$. Similarly, if $n=3$, the $8$ nodes on level $0$ and the $4$ nodes on level $1$ end up with a balance of $-7$ each, while the $2$ nodes on level $2$ and the root node end up with a balance of $28$ each.
Now assume that $n\ge 4$. Then each of the $2^n$ nodes on level $0$ and the $2^{n-1}$ nodes on level $1$ pays $7$ and receives nothing for a balance of $-7$. If $2\le\ell\le n-2$, each node on level $\ell$ receives $4\cdot7=28$ from its four grandchildren and pays $7$ to its grandparent, so it ends up with a balance of $21$. Finally, each node on levels $n$ and $n-1$ receives $4\cdot7=28$ from its grandchildren and pays nothing, ending with a balance of $28$.
As a quick partial check, note that there are $2^{n-\ell}$ nodes on level $\ell$, so for $n\ge 4$ the sum of the final balances of all nodes is
$$\begin{align*}
-7\left(2^n+2^{n-1}\right)+21\sum_{\ell=2}^{n-2}2^{n-\ell}+28(2+1)&=-7\left(2^n+2^{n-1}\right)+21\sum_{k=2}^{n-2}2^k+3\cdot28\\
&=-21\cdot2^{n-1}+21\left(2^{n-1}-4\right)+4\cdot21\\
&=0\;,
\end{align*}$$
just as it should be: there is no net flow of money into or out of the system.
