$\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$As shown in the post by Alexandre Eremenko, the answer to this question is yes if $C=1$ (and hence if $C\le1$).
On the other hand, fedja's example shows that the answer is no if $C\ge e$.
Here it will be shown, by a slight modification of fedja's example, that the answer will remain no for any $C>1$. It will also be shown that the functions $f_n$ can be taken to be (some explicitly constructed) polynomials (in fedja's example, the $f_n$'s are scalar multiples of products of Blaschke factors).
Specifically, the following will be shown:
For each real $C>1$, there is a sequence $(f_n)$ of polynomials over $\R$ such that for all natural $n$ and $m$ we have
\begin{equation*}
f_n(0)=1\quad\text{and}\quad |f_n^{(m)}(0)|\le C^n m!, \tag{1}\label{1}
\end{equation*}
whereas for each $\de\in(0,1)$
\begin{equation*}
\liminf _n |E_{n,\de}|>0, \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
E_{n,\de}:=\{x\in[0,\de]\colon |f_n(x)|\le e^{-n}\}.
\end{equation*}
To prove this, take indeed any real $C>1$. Then there exist sequences $(x_j)$ and $(p_j)$ in $(0,1)$ such that $x_j\downarrow0$, $p_j\downarrow0$,
\begin{equation*}
c:=\prod_j x_j^{p_j}=\exp\sum_j p_j \ln x_j\in\Big(\frac1{\sqrt C},1\Big), \quad\text{and}\quad
K:=\exp\sum_j p_j x_j<\sqrt C;
\end{equation*}
for instance, one may let $x_j:=1/j$ and $p_j:=\ep/j^2$ for a small enough $\ep>0$ and all $j$.
For natural $n$ and all complex $z$, let then
\begin{equation*}
g_n(z):=\prod_j(x_j-z)^{\lfloor np_j\rfloor} \quad\text{and}\quad
f_n(z):=\frac{g_n(z)}{c_n},
\quad\text{where}\quad c_n:=g_n(0)=\prod_j x_j^{\lfloor np_j\rfloor}.
\end{equation*}
Since $p_j\downarrow0$, (for each $n$) the function $g_n$ is a polynomial and hence $f_n$ is a polynomial. Note also that $f_n(0)=1$ and $c_n\ge c^n$. Moreover, for all complex $z$ with $|z|\le1$,
\begin{equation*}
|g_n(z)|\le\prod_j(x_j+1)^{np_j}\le\exp\sum_j np_j x_j=K^n
\end{equation*}
and hence
\begin{equation*}
|f_n(z)|\le\frac{K^n}{c_n}\le\frac{K^n}{c^n}<C^n.
\end{equation*}
So, in view of Cauchy's differentiation formula, we have \eqref{1}.
Now take any $\de\in(0,1)$. In view of the conditions $p_j\downarrow0$ and $c\in(\frac1{\sqrt C},1)$, we have
\begin{equation}
h_j:=\Big(\frac c{2e}\Big)^{1/p_j}\downarrow0
\end{equation}
(as $j\to\infty$).
Recalling also that $x_j\downarrow0$, we see that there is some natural $k=k_\de$ such that $(x_k,x_k+h_k)\subseteq(0,\de)$ and for all $x\in(x_k,x_k+h_k)$
\begin{equation}
|f_n(x)|\le\frac1{c_n}\,h_k^{np_k-1}\prod_{j\colon j\ne k}1
\le\frac1{c^n}\,h_k^{np_k-1}
=\frac1{h_k(2e)^n}\le e^{-n}
\end{equation}
eventually (that is, for all large enough $n$). So, eventually $(x_k,x_k+h_k)\subseteq E_{n,\de}$, which proves that \eqref{2} holds as well. $\quad\Box$
