Flatness of the completion is Algebra, Lemma \ref{algebra-lemma-completion-flat}. To show that the map is formally unramified, it suffices to show that $\Omega_{A^\wedge/A} = 0$, see Algebra, Lemma \ref{algebra-lemma-characterize-formally-unramified}. \medskip\noindent We sketch a proof. Choose $x_1, \ldots, x_r \in A$ which map to a $p$-basis $\overline{x}_1, \ldots, \overline{x}_r$ of $\kappa$, i.e., such that $\kappa$ is minimally generated by $\overline{x}_i$ over $\kappa^p$. Choose a minimal set of generators $y_1, \ldots, y_s$ of $\mathfrak m$. For each $n$ the elements $x_1, \ldots, x_r, y_1, \ldots, y_s$ generate $A/\mathfrak m^n$ over $(A/\mathfrak m^n)^p$ by Frobenius. Some details omitted. We conclude that $F : A^\wedge \to A^\wedge$ is finite. Hence $\Omega_{A^\wedge/A}$ is a finite $A^\wedge$-module. On the other hand, for any $a \in A^\wedge$ and $n$ we can find $a_0 \in A$ such that $a - a_0 \in \mathfrak m^nA^\wedge$. We conclude that $\text{d}(a) \in \bigcap \mathfrak m^n \Omega_{A^\wedge/A}$ which implies that $\text{d}(a)$ is zero by Algebra, Lemma \ref{algebra-lemma-intersect-powers-ideal-module-zero}. Thus $\Omega_{A^\wedge/A} = 0$. \medskip\noindent Suppose $A$ is regular. Then, using the Cohen structure theorem $x_1, \ldots, x_r, y_1, \ldots, y_s$ is a $p$-basis for the ring $A^\wedge$, i.e., we have $$ A^\wedge = \bigoplus\nolimits_{I, J} (A^\wedge)^p x_1^{i_1} \ldots x_r^{i_r} y_1^{j_1} \ldots y_s^{j_s} $$ with $I = (i_1, \ldots, i_r)$, $J = (j_1, \ldots, j_s)$ and $0 \leq i_a, j_b \leq p - 1$. Details omitted. In particular, we see that $\Omega_{A^\wedge}$ is a free $A^\wedge$-module with basis $\text{d}(x_1), \ldots, \text{d}(x_r), \text{d}(y_1), \ldots, \text{d}(y_s)$. \medskip\noindent Now if $A \to A^\wedge$ is formally \'etale or even just formally smooth, then we see that $\NL_{A^\wedge/A}$ has vanishing cohomology in degrees $-1, 0$ by Algebra, Proposition \ref{algebra-proposition-characterize-formally-smooth}. It follows from the Jacobi-Zariski sequence (Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL}) for the ring maps $\mathbf{F}_p \to A \to A^\wedge$ that we get an isomorphism $\Omega_A \otimes_A A^\wedge \cong \Omega_{A^\wedge}$. Hence we find that $\Omega_A$ is free on $\text{d}(x_1), \ldots, \text{d}(x_r), \text{d}(y_1), \ldots, \text{d}(y_s)$. Looking at fraction fields and using that $A$ is normal we conclude that $a \in A$ is a $p$th power if and only if its image in $A^\wedge$ is a $p$th power (details omitted; use Algebra, Lemma \ref{algebra-lemma-derivative-zero-pth-power}). A second consequence is that the operators $\partial/\partial x_a$ and $\partial/\partial y_b$ are defined on $A$. \medskip\noindent We will show that the above lead to the conclusion that $A$ is finite over $A^p$ with $p$-basis $x_1, \ldots, x_r, y_1, \ldots, y_s$. This will contradict the non-excellency of $A$ by a result of Kunz, see \cite[Corollary 2.6]{Kun76}. Namely, say $a \in A$ and write $$ a =  \sum\nolimits_{I, J} (a_{I, J})^p x_1^{i_1} \ldots x_r^{i_r} y_1^{j_1} \ldots y_s^{j_s} $$ with $a_{I, J} \in A^\wedge$. To finish the proof it suffices to show that $a_{I, J} \in A$. Applying the operator $$ (\partial/\partial x_1)^{p - 1} \ldots (\partial/\partial x_r)^{p - 1} (\partial/\partial y_1)^{p - 1} \ldots (\partial/\partial y_s)^{p - 1} $$ to both sides we conclude that $a_{I, J}^p \in A$ where $I = (p - 1, \ldots, p - 1)$ and $J = (p - 1, \ldots, p - 1)$. By our remark above, this also implies $a_{I, J} \in A$. After replacing $a$ by $a' = a - a_{I, J}^p x^I y^J$ we can use a $1$-order lower differential operators to get another coefficient $a_{I, J}$ to be in $A$. Etc.