$\newcommand{\R}{\mathbb{R}}
\newcommand{\si}{\sigma}
\newcommand{\supp}{\operatorname{\mathrm supp}}
\newcommand{\cch}{\operatorname{\mathrm cch}}
$
If $F\in L^2$, then the condition
\begin{equation*}
|\mu_n(F)|(\ln n)^n<C<\infty\quad\forall n\in\mathbb{N} \tag{1}
\end{equation*}
implies that $F=0$ almost everywhere (a.e.) on $[0,1]$.

Indeed, let
\begin{equation*}
f(z):=\sum_{n=0}^\infty\frac{\mu_n(F)(iz)^n}{n!}
=\int_0^1 F(t)\,dt \sum_{n=0}^\infty\frac{(itz)^n}{n!}
=\int_0^1 F(t)e^{itz}\,dt
\end{equation*}
for all complex $z$.
By (1), for any natural $k\ge2$
\begin{equation*}
|f(z)|\le O(1+|z|^{k-1})+C\sum_{n=k}^\infty\frac{(|z|/\ln n)^n}{n!}
\le O(1+|z|^{k-1})+C\sum_{n=0}^\infty\frac{(|z|/\ln k)^n}{n!}
\le c_k e^{|z|/\ln k}
\end{equation*}
for some real $c_k>0$ and all complex $z$.
So, $f(z)$ is an entire function of exponential type $a$ for any real $a>0$.

Hence, by a Paley--Wiener theorem (more specifically, see e.g. Theorem 19.3 on page 375), for each real $a>0$ there is an $L^2$ function $F_a$ such that for all complex $z$
\begin{equation*}
f(z)=\int_{-a}^a F_a(t)e^{itz}\,dt
\end{equation*}
Taking the inverse Fourier transform, we see that for each $a\in(0,1)$ and all complex $z$
\begin{equation*}
f(z)=\int_0^a F(t)e^{itz}\,dt.
\end{equation*}
Thus, for all complex $z$
\begin{equation*}
0=f(z)=\int_0^1 F(t)e^{itz}\,dt.
\end{equation*}
So, indeed $F=0$ a.e. on $[0,1]$.

Consider now the more general setting when $F(t)\,dt$ is replaced by $\rho(dt)$, where $\rho$ is a signed measure over $\R$ with support $\supp\rho\subseteq[0,1]$. Then condition (1) is replaced by
\begin{equation*}
|\mu_n(\si)|(\ln(n+2))^n<C<\infty\quad\forall n\in\mathbb{N}_0:=\{0,1,\dots\}, \tag{1a}
\end{equation*}
where
\begin{equation*}
\mu_n(\si):=\int_0^1 t^n\si(dt)=\int_0^1 t^{n+2}\rho(dt)\quad\text{and}\quad\si(dt):=t^2\rho(dt).
\end{equation*}
So, $\supp\si\subseteq[0,1]$. Also, (1a) implies $\mu_0(\si)<\infty$, that is, the signed measure $\si$ is finite. So, by reasoning quite similar to that in the above $L^2$ case, we see that under condition (1a)
\begin{equation*}
f(z):=\int_\R e^{itz}\,\si(dt)
\end{equation*}
is an entire function of exponential type $a$ for any real $a>0$.

For real $b>0$ and $t\in\R$, let
\begin{equation*}
G_b(t):=(g_b*d\si)(t):=\int_\R g_b(t-s)\,\si(ds)
=\frac{\si([t-b,t+b])}{2b},
\end{equation*}
where $g_b:=\frac1{2b}\,1_{[-b,b]}$. Then $G_b\in L^2(\R)$, since the signed measure $\si$ is finite with $\supp\si\subseteq[0,1]$. Also,
\begin{equation*}
\hat g_b(z):=\int_\R e^{itz}g_b(t)\,dt=\frac{\sin bz}{bz}
\end{equation*}
for $z\ne0$, and hence
\begin{equation*}
f_b(z):=\int_\R e^{itz}\,G_b(t)\,dt
=\int_\R e^{itz}\,(g_b*d\si)(t)\,dt=\hat g_b(z)f(z)
\end{equation*}
is of exponential type $b+a$ for all real $a>0$. Thus, by the cited Paley--Wiener theorem, $\supp(g_b*d\si)\subseteq[-b-a,b+a]$ for all real $a>0$ and hence
\begin{equation*}
\supp(g_b*d\si)\subseteq[-b,b].
\end{equation*}

On the other hand, because $\supp g_b$ and $\supp\si$ are both compact, by Theorem 4.3.3 on page 117, $\cch\supp(g_b*d\si)=\cch\supp g_b+\cch\supp\si$, where $\cch$ denotes the closed convex hull. Since $\cch\supp g_b\ni0$, we conclude that $\cch\supp\si\subseteq\cch\supp(g_b*d\si)\subseteq[-b,b]$, whence
$\supp\rho\subseteq\supp\si\cup\{0\}\subseteq[-b,b]$, for any real $b>0$. Thus, condition (1a) implies
\begin{equation*}
\supp\rho\subseteq\{0\}. \tag{2}
\end{equation*}

Vice versa, trivially (2) implies (1a). Thus, (1a) holds iff $\supp\rho\subseteq\{0\}$.

**Response to the second comment by the OP:**
Let us try to sort all this out.

Your condition
\begin{equation}
0<c<|\mu_n(F)|(\ln n)^n<C<\infty \quad\forall n\in\mathbb{N} \tag{3}
\end{equation}
can actually never hold as stated, because the inequalities $0<c<|\mu_n(F)|(\ln n)^n$ will always be false for $n=1$. So, of course, I assumed (3) with $\mathbb{N}$ replaced by $\{2,3,\dots\}$.

In your post here, you did not even mention "analytical distributions" or Ehrenpreis. Instead, you wrote: "Let $F:[0,1]\to\mathbb{R}$ be a measurable function" and then you also wrote "In fact, I do not need $F$ be a function; $F(x)dx$ may be a signed measure on $[0,1]$".

If $F(x)dx$ is replaced by a signed measure $\rho(dx)$, then the corrected version of your condition (3) can be rewritten as the conjunction of my condition (1a) and the condition
\begin{equation}
0<c<|\mu_n(\si)|(\ln n)^n \quad\forall n\in\{2,3,\dots\}. \tag{4}
\end{equation}

It was shown in this answer that, under condition (1a), the support of the signed measure $\rho$ must be contained in the set $\{0\}$. (Method-wise, this was first done in the case when $\rho(dx)=F(x)dx$ with $F\in L^2$, and then extended to the general case of any signed measure $\rho$ on $[0,1]$.)

It then immediately follows that there is no signed measure $\rho$ on $[0,1]$ satisfying (1a) and (4) -- or, equivalently, satisfying the corrected version of your condition (3). In other words, your conditions do result in a contradiction. This completely answers your posted question.