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).