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.