The answer by Bazin (https://mathoverflow.net/users/21907/bazin), Faa di Bruno's formula for vector valued functions, URL (version: 2012-09-04): https://mathoverflow.net/q/106339
is providing a formula for the $n$th derivative of $f\circ g$. Assuming that $g=\Phi$ is invertible and smooth, we take $f=\Phi^{-1}$ its inverse function and we find for $n\ge 2$
$$
0=\sum_{\substack{n_1+\dots+n_r=n\\r\ge 1, n_j\ge 1}}
\frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!},
$$
so that writing the "first" term for $r=n, n_j=1$, we get
$$
\frac{f^{(n)}\circ g}{n!}\bigl(g^{(1)}\bigr)^n+\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}}
\frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!}=0,
\tag{1}$$
and thus an explicit formula expressing $$
\bigl[(\Phi^{-1})^{(n)}\circ \Phi\bigr]\times \frac{(\Phi^{(1)})^n}{n!}
$$
as a linear combination of the $r$th derivative of $\Phi^{-1}$ (with $1\le r<n$)
whose coefficients are products $\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}$ where $n_1+\dots+n_r=n$.
We get that
$$
\underbrace{[(\Phi^{-1})^{(n)}\circ \Phi]\times \frac{(\Phi^{(1)})^n}{n!}}_{\Psi_n}
=-\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}}
\underbrace{{[(\Phi^{-1})^{(r)}\circ \Phi]}
\times \frac{(\Phi^{(1)})^r}{r!}}_{=\Psi_r}
{(\Phi^{(1)})^{-r}}
\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}
,
$$
that is
$$
\Psi_n=-\sum_{n>r\ge 1}
\Psi_r
\underbrace{
{(\Phi^{(1)})^{-r}}
\sum_{\substack{n_1+\dots+n_r=n\\ n_j\ge 1}}
\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}}_{\Omega_{r,n}(\Phi)},\quad
\text{and we have also }
\Psi_1=I,
$$
so that
$$
\Psi_n=-\sum_{1\le r\le n-1}\Psi_r \Omega_{r,n}(\Phi), \quad \Psi_1=
I.\tag{$\ast$}$$
We can now prove inductively that $\Vert\Psi_n\Vert$ is bounded above by a polynomial $$
P_n(\frac{\Vert \Phi^{(\nu_1)}\Vert}{\Vert \Phi^{(1)}\Vert },
\dots,
\frac{\Vert \Phi^{(\nu_l)}\Vert}{\Vert \Phi^{(1)}\Vert }), \quad \nu_j\le n.
$$
To actually get a bound,
I have indeed to clarify the meaning of the above formulas in a multidimensional setting.
We stay with a smooth invertible mapping
$$
\Phi:U\longrightarrow V, \quad U,V \text{open subsets of $X,Y$, Banach spaces,}
$$
and we note $F$ the inverse mapping of $\Phi$,
$$x\in U, S\in X, y=\Phi(x), T=\Phi'(x) S\in Y.
$$
We recall that $\Phi^{(n)}(x)$ is a $n$-multilinear symmetric form
on $X$, valued in $Y$.
Applying (1), we obtain for $n\ge 2$,
\begin{multline}
\frac{F^{(n)}(\Phi(x))}{n!}\bigl(\Phi'(x) S\bigr)^n
\\+\sum_{\substack{n_1+\dots+n_r=n\\1\le r\le n-1, n_j\ge 1}}
\frac{F^{(r)}(\Phi(x))}{r!}
\left(\frac{\Phi^{(n_1)}(x)S^{n_1}}{n_1!},\dots,\frac{\Phi^{(n_r)}(x)S^{n_r}}{n_r!}\right)=0,
\tag{2}
\end{multline}
so that
\begin{multline}
\frac{F^{(n)}(y)}{n!}\bigl(T\bigr)^n
\\+\sum_{\substack{n_1+\dots+n_r=n\\1\le r\le n-1, n_j\ge 1}}
\frac{F^{(r)}(y)}{r!}\left(\frac{\Phi^{(n_1)}(x)[F'(y) T]^{n_1}}{n_1!},\dots,\frac{\Phi^{(n_r)}(x)[F'(y) S]^{n_r}}{n_r!}\right)=0.
\tag{3}
\end{multline}
As a consequence, we obtain that for $n\ge 2$,
\begin{multline}
\frac{\Vert F^{(n)}(y)\Vert}{n!}\le
\\\sum_{\substack{n_1+\dots+n_r=n\\1\le r\le n-1, n_j\ge 1}}
\frac{\Vert F^{(r) }(y)\Vert}{r!}\frac{\Vert \Phi^{(n_1)}(x)\Vert \Vert F'(y) \Vert ^{n_1}}{n_1!}\dots \frac{\Vert \Phi^{(n_r)}(x)\Vert \Vert F'(y) \Vert ^{n_r}}{n_r!}
\\
=\Vert F'(y) \Vert ^{n}
\sum_{\substack{n_1+\dots+n_r=n\\1\le r\le n-1, n_j\ge 1}}
\frac{\Vert F^{(r) }(y)\Vert}{r!}\frac{\Vert \Phi^{(n_1)}(x)\Vert }{n_1!}\dots \frac{\Vert \Phi^{(n_r)}(x)\Vert }{n_r!}.
\tag{4}
\end{multline}
Using that formula, you get easily
$$
\Vert F^{(2)}(y)\Vert\lesssim
\Vert F^{(1)}(y)\Vert^{3}\Vert \Phi^{(2)}(x)\Vert,
$$
as well as
$$
\Vert F^{(3)}(y)\Vert\lesssim
\Vert F^{(1)}(y)\Vert^{4}\Vert \Phi^{(3)}(x)\Vert
+\Vert F^{(1)}(y)\Vert^{6}
\Vert \Phi^{(1)}(x)\Vert
\Vert \Phi^{(2)}(x)\Vert^2.
$$
The last formula indicates some differences with the one-dimensional case:
we have indeed
$
F'(y)\Phi'(x)=I
$
which implies
$$
1\le\Vert F'(\Phi(x))\Vert\Vert\Phi'(x)\Vert,
$$
but not a bound from above for the rhs. However, you can say that, for $x$ in a neighborhood of $x_0$
$$
1\le\Vert F'(\Phi(x))\Vert\Vert\Phi'(x)\Vert\le 2
\Vert F'(\Phi(x_0))\Vert\Vert\Phi'(x_0)\Vert,
$$
and get for instance
$$
\Vert F^{(3)}(y)\Vert\lesssim
\Vert F^{(1)}(y)\Vert^{5}\bigl(
\Vert \Phi^{(1)}(x)\Vert\Vert \Phi^{(3)}(x)\Vert
+
\Vert \Phi^{(2)}(x)\Vert^2
\bigr),
$$
which corresponds to your claim.
Except for that multidimensional difficulty,
the induction proof seems to work from Formula (4):
using the latter, we can prove for $n\ge 2$,
\begin{equation}
\Vert F^{(n)}(y)\Vert\lesssim
\Vert F^{(1)}(y)\Vert^{2n-1}\sum_{\substack{
\nu_1+\dots+\nu_{n-1}=2n-2
}}\prod_{1\le j\le n-1}\Vert \Phi^{\nu_j}(x)\Vert.
\tag{5}\end{equation}
