Does the series $\sum_{i=0}^{\infty}\exp\{{-(m!)!}\}(D^m\phi)(0)$ converge for every $\phi \in C^\infty$? 
Does the series $$\sum_{m=0}^{\infty}\exp\{{-(m!)!}\}(D^m\phi)(0)$$
  converge for every $\phi \in C^\infty$?

For analytic function $\phi$, we can show that the series converges by using Caushy-Schwartz inequality. But I believe in general that there is an example that the series diverges.
Although we have a Taylor expansion
$$\phi(x)-\phi(0)=x\phi'(x)+...+\frac{x^n}{n!}\phi^{(n)}(x)+\int_0^x\frac{y^n}{n!}\phi^{(n)}(y)dy$$
it seems to be useless because we don't have any information of $\phi^{(n)}(y)$.
How to make the growth of $\phi^{(n)}(0)$ fast?
 A: One way to assure the convergence of your series is to assume that the sequence $b_m =(D^m\phi)(0) $ to be bounded sequence since you have no problem with the convergence of the series $ \sum_{m=0}^{\infty}\exp\{{-(m!)!}\} $. This is due to the fact: 

If $|b_n| \leq M$ and $\sum_{n} a_n < \infty$ then $\sum_{n}a_nb_n$ converges. 

A: Choose $\psi \in C^{\infty}(\mathbb{R})$ so that $ \mathbf{1}_{[-1,1]} \leq \psi \leq \mathbf{1}_{[-2,2]} $ on $\mathbb{R}$. Then let
$$ \phi(x) = \sum_{m=0}^{\infty} e^{(m!)!} \frac{x^m}{m!} \psi(a_n x), $$
where $(a_m)_{m\geq 0}$ is a sequence of positive real numbers chosen such that $a_m \geq 1$ for all $m$ and
$$ \sum_{m=0}^{\infty} e^{(m!)!}(2/a_m)^m < \infty. $$
Writing $\|f\| = \sup_{x\in\mathbb{R}} |f(x)|$ for the supremum norm on $\mathbb{R}$, for each $ k \geq 0$, we get
\begin{align*}
&\sum_{m=0}^{\infty} e^{(m!)!} \left\| D^k\left(\frac{x^m}{m!} \psi(a_n x)\right) \right\| \\
&= \sum_{m=0}^{\infty} e^{(m!)!} \left\| \sum_{j=0}^{k} \binom{k}{j} \frac{x^{m-j}}{(m-j)!} a_m^{k-j}\psi^{(k-j)}(a_m x) \right\| \\
&\leq \sum_{m=0}^{\infty} e^{(m!)!} \sum_{j=0}^{k} \binom{k}{j} (2/a_m)^{m-j} a_m^{k-j} \bigl\| \psi^{(k-j)} \bigr\| \\
&\leq \Biggl[ \sum_{m=0}^{\infty} e^{(m!)!} 2^{m+k} a_m^{k-m} \Biggr] \max_{0\leq i\leq k} \bigl\| \psi^{(i)} \bigr\| \\
& < \infty.
\end{align*}
So, $\phi$ is a smooth function on $\mathbb{R}$ and term-wise differentiation is applicable indefinitely. However,
$$ D^k\phi(0) = \sum_{m=0}^{\infty} e^{(m!)!} \left. D^k\left(\frac{x^m}{m!} \psi(a_n x)\right) \right|_{x=0} = e^{(k!)!}, $$
and so, the sum in question diverges.
