Let's pursue Jochen's idea. We assume $A \ne \emptyset.$ 

Let 
$$ \varphi(t) = \begin{cases}  
   e^{-\frac{1}{t}} &\text{if $ t>0$}\\
   0                &\text{otherwise.}
\end{cases}$$ 

This function is $\mathcal C^{\infty}$, and $0 < \varphi(t)$ iff $0<t.$ 

Define $\rho$ as 
$$\rho(x) := k \varphi(1- \|x\|_2^2)$$ 
where $k$ is chosen so that $\int \rho(x)\space dx =1.$
$\rho$ is  $\mathcal C^{\infty}$, $0 \le \rho$, and $0< \rho(x)$ iff $\|x\|_2^2 < 1.$  

Let $\delta > 0$. Set $\rho_{\delta}(x) = \frac{1}{\delta^n} \rho(\frac{x}{\delta})$. Note that $0<\rho_{\delta}(x)$ iff $ \|x\|_p < \delta$, and $\int \rho_\delta = 1.$ 


Let $\mathscr O = A+B_2(0,\delta)$, and consider the function  
$$  f =   I_{\mathscr O} \ast \rho_{\delta} .$$ 

 
Clearly $0 \le f \le 1$ and $f$ is $\mathcal C^{\infty}.$

If $a \in A$, then $B_2(a,\delta) \subset A+B_2(0,\delta) = \mathscr O $, so
$ I_{\mathscr O } \ast \rho_{\delta}(a) = 1.$   

If $x \notin A+B_2(0,\delta)$ then 
$B_2(x,\delta) \cap (A + B_2(x,\delta)) = emptyset$, 
so 
$ I_{\mathscr O } \ast \rho_{\delta}(x) = 0.$


For derivatives of any order:  
$$ D^\nu f(x) = 
D^\nu \int_{\mathscr O} \frac{1}{\delta^n} \rho(\frac{x-y}{\delta}) dy = \frac{1}{\delta ^{|\nu|}} \int_{\mathscr O} \frac{1}{\delta^n} (D^\nu\rho)(\frac{x-y}{\delta}) dy
$$
so
$$ |D^\nu f(x)| \le 
( \int |(D^\nu\rho)(y)| dy)\frac{1}{\delta ^{|\nu|}}.
$$


Finally, for any $p$, there is a constant C (depending on $p$ and $n$) such that $B_2(0,1) \subset B_p(0,C_p)$. Therefore:  
$$ A \subset A+B_2(0,\delta) \subset A+B_2(0,2\delta) \subset A+B_p(0,2C_p\delta) = A + B_p(0,\epsilon) = A^{\epsilon} $$
if we take $\delta = \epsilon /(2C_p) .$ 

With this choice of $\delta$, the function $f$ satisfies all the requirements.


$$\dots\dots\dots$$
In order to apply the result without distinguishing the dimention $n$ of the underlying space, it is of interest to consider the following lemma.

$\mathbf{Lemma}$: Given $ p \in \{1,2\} \cup (3,+\infty]$, there is $\rho:\mathbb R^n \rightarrow \mathbb R$ such that:  
i) $\rho$ is $\mathscr C^3$   
ii) $0 \le \rho$   
iii) $\rho(x) > 0$ iff $\|x\|_p < 1$  
(iv) and, by re-scaling,  $\int \rho(x) \space dx = 1$) 

With this choice of $\rho$ the original argument can be carried without resorting to the 2-norm.

$\mathbf{Proof}$: With $\varphi$ as above, if $p \in \{2\} \cup (3,+\infty)$, define $\rho$ as
$$\rho(x) := \varphi(1- \|x\|_p^p) .$$
With this choice, $\rho$ satisfies i,ii,iii. The condition $p>3$ is needed to justify the claimed smoothness of $\rho.$   

If $p=1$ or $p=\infty$, note that the open unit ball in $(\mathbb R^n , \|.\|_p)$ is a polytope of the form $P=\{x| v_i^{'}x<1\}.$ When $p=\infty$, $\{v_i\} = \{\pm e_j\}$ where $\{e_j\}$ is the canonical (vector space) bases of $\mathbb R^n$. When $p=1$, $\{v_i\} = \{-1,1\}^n$. In these cases, take $\rho$ as
$$ \rho(x) = \prod_i \space \varphi(1-v_i^{'}x) . \blacksquare$$