A special type of generating function for Fibonacci Notation. Let $[x^n]G(x)$ be the coefficient of $x^n$ in the Taylor series of $G(x)$.
Consider the sequence of central binomial coefficients $\binom{2n}n$. Then there two ways to recover them:
$$\binom{2n}n=[x^n]\left(\frac1{\sqrt{1-4x}}\right) \tag{1-1}$$
and
$$\binom{2n}n=[x^n]\left((1+x)^2\right)^n. \tag{1-2}$$
This time, take the Fibonacci numbers $F_n$ then, similar to (1-1), we have
$$F_n=[x^n]\left(\frac1{1-x-x^2}\right). \tag{2-1}$$
I would like to ask:

QUESTION. Does there exist a function $F(x)$, similar to (1-2), such that
  $$F_n=[x^n]\left(F(x)\right)^n? \tag{2-2}$$

 A: The power series $F(x)$ is closely related to the series of the "exponential reversion of Fibonacci numbers" $$R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}$$  (the $r_n$ are A258943, quoted in a comment). In fact it appears that, again in the notation of Henri Cohen, $$a_{n+1}=nr_n,$$ equivalently $$F'(x)=xR'(x).$$
So if the Fibonacci numbers are encapsulated by $$x=\sum_{n\ge1}F_n\frac{y^n}{n!}=y+\frac{y^2}{2!}+2\frac{y^3}{3!}+3\frac{y^4}{4!}+5\frac{y^5}{5!}+\cdots,$$ the reverse series of this is 
$$ 
y=R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}=x-\frac{x^2}{2!}+\frac{x^3}{3!}+\color{red}{2}\frac{x^4}{4!}-\color{red}{25}\frac{x^5}{5!}+-\cdots,$$
while $$\begin{align}F(x)=1+\sum_{n\ge1}a_n {x^n} &=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+8\frac{x^5}{5!}-125\frac{x^6}{6!}+-\cdots\\
&=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+4\cdot\color{red}{2}\frac{x^5}{5!}-5\cdot\color{red}{25}\frac{x^6}{6!}+-\cdots\end{align}.$$
Possibly this relationship is not even specific to the Fibonacci numbers.  
EDIT: It looks like the sequence $\{a_n\}$ has finally yielded its secrets. Given the conjectured "smoothness" of these coefficients (i.e. all prime factors are relatively small) as mentioned in Henri Cohen's answer, I have looked again into the factors and those quadratic sequences mentioned in the comments, and fortunately there are enough primes in them, such that finally I was able to find the pattern! We have for the sequence $\{r_n\}$ $${ r_n=\begin{cases} {(-1)^k} \prod\limits_{j=1}^k(n^2-5nj+5j^2) \quad\text{for 
 }\ n=2k-1, \\ \\  {(-1)^kk\cdot} \prod\limits_{j=1}^{k-1}(n^2-5nj+5j^2) \quad\text{for  }\ n=2k. \end{cases}}$$ Once found, it should not be hard to prove that rigorously.
As pointed out by Agno in a comment, we can reduce to linear factors and write the product in terms of the Gamma function simply as $$r_n= {\sqrt5^{ \,n-1 }\frac { \Gamma \left( \frac{5-\sqrt {5}}{10}n \right)}{  \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) }}.$$More generally, if we start with a Lucas sequence $$f_0=0,\ f_1=1,\ f_n=pf_{n-2}+qf_{n-1}\quad(n\ge2),$$ the reversed series has  $$\boxed{r_n= {\sqrt{4p+q^2}^{ \,n-1 }\frac { \Gamma \left[\dfrac n2 \Bigl(1-\dfrac{q}{\sqrt{4p+q^2}}  \Bigr)\right]}{ \Gamma \left[1-\dfrac n2 \Bigl(1+\dfrac{q}{\sqrt{4p+q^2}}  \Bigr)\right]}}}.$$ Note that whenever the argument in the denominator is a negative integer, the coefficient $r_n$ vanishes, e.g. this happens when $p=3,q=2$ for all $n\equiv0\pmod4$.  
As far as the sequence of the signs, it is after all quite regular and is in fact self-similar (that is, unless $\sqrt{4p+q^2}$ is rational). This self-similar behaviour of the signs can be seen by virtue of the (negative) argument of the Gamma function in the denominator, knowing that $\Gamma$ changes signs at each negative integer and the multiples of $\sqrt{4p+q^2}$ occurring in the argument do the rest. (Think e.g. of the self similarity features of the Wythoff sequence.)
For the reversion of the original Fibonacci sequence, I have displayed here the signs of the first $1500$ even and then the first $1500$ odd coefficients and found that their quasi periodicity comes out nicely when putting exactly $76$ in each row (writing "o" instead of "$-$" for better visibility). The "longest pairings" are colored:
all patterns "$++--++--++$" in yellow and
all patterns "$--++--++--$" in blue. 
A: I started writing this before Richard's answer appeared, with which it overlaps a lot, but I still have something to add.
Let us look at a more general problem: Suppose that $G(x) = 1+g_1x+g_2x^2+\cdots$ and that 
$[x^m]G(x)^m = c_m$ for $m\ge1$. It is clear that the $g_i$ can be expressed uniquely in terms of the $c_i$, and we would like to find an explicit formula. 
Let 
$$
R(x) = \exp\biggl(-\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr)
$$ 
and let $f=f(x)$ satisfy $f = xR(f)$, so $f(x) =\bigl(x/R(x)\bigr)^{\langle -1\rangle}$.
By formula (2.2.7) of my survey paper on Lagrange inversion (a corollary of Lagrange inversion), 
$[x^m](x/f)^m = c_m$, so by the uniqueness of $G(x)$, we have $G(x) = x/f(x)$. By formula (2.2.4) of my paper (a form of Lagrange inversion) we have for $\alpha\ne -m$,
$$[x^m] (f/x)^{\alpha}=\frac{\alpha}{m+\alpha} [x^m] R(x)^{m+\alpha}$$
so
$$
[x^m] G(x)^{-\alpha}=\frac{\alpha}{m+\alpha} [x^m] \exp\biggl(-(m+\alpha)\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr).
$$
In particular, taking $\alpha=-1$ gives the formula for the coefficients of $G(x)$ in terms of the $c_i$:
we have $g_1=c_1$ and for $m>1$,
$$
g_m=-\frac{1}{m-1} [x^m] \exp\biggl(-(m-1)\sum_{n=1}^\infty \frac{c_n}{n} x^n\biggr).
$$
We can use these formulas to find some nice examples of $[x^m]G(x)^m=c_m$. 
First take $c_m$ to be the constant $C$. Then $R(x)=\exp(-\sum_{n=1}^\infty C x^n/n) = (1-x)^C$, $f$ satisfies $f=x(1-f)^C$, and 
$$
\begin{aligned}G(x)^{-\alpha} &= \sum_{m=0}^\infty  (-1)^m\frac{\alpha}{m+\alpha} \binom{C(m+\alpha)}{m} x^m\\
&=1+\sum_{m=1}^\infty (-1)^m \frac{\alpha C}{m}\binom{C(m+\alpha)-1}{m-1}x^m.
\end{aligned}
$$
In particular, if $C=1$ then $f=x/(1+x)$ and $G=1+x$. If $C=-1$ then $f$ is $xc(x)$ where $c(x)$ is the Catalan number generating function, $c(x) =(1-\sqrt{1-4x})/(2x)$,  and $G(x) = 1/c(x)$. If $C=2$ then $f=xc(-x)^2$, so $G(x) = c(-x)^{-2}$.
For another example take $c_1=-1$ and  $c_m=0$ for $m>1$. Then $R(x) = e^{x}$ so $f(x)$ is the "tree function" satisfying $f = xe^f$, $G = x/f = e^{-f}$ and 
$$G(x)^{-\alpha} = \sum_{m=0}^\infty \alpha (m+\alpha)^{m-1}\frac{x^m}{m!}.$$
A: For any desired sequence $a_0,a_1,a_2,\cdots$ there is a unique function (a formal power series) $F(x)$ with $[x^n]F(x)=a_n$ namely $F(x)=\sum a_ix^i.$
Sometimes $F(x)$ is a polynomial. This happens exactly when the sequence is zero from some point on. Sometimes $F(x)$ is a rational function. This happens exactly when the sequence satisfies a linear homogeneous recurrence relation with constant coefficients. Then the denominator is determined by the recurrence and the numerator by the initial conditions. And $F(x)$ might or might not be expressible in a closed form of some other given type, such as $\frac{P(x)}{\sqrt[k]{Q(x)}}$ for $P,Q$ polynomials.

You desire that there is a function $G(x)$ with $[x^n](G(x)^n)=a_n$ for all $n$.  Again there is a unique formal power series $G[x]=1+s_1x+s_2x^2+\cdots$ such that $$[x^n](G(x)^n)={\large \lbrace}\begin{array}{lr}
        1 & \text{for } n=0\\
        a_n & \text{for } n \geq 1\\
        \end{array}    $$ 
So you need to either have $a_0=1$ or restrict the requirement to $n \geq 1.$
The unique formal power series $G(x)$ is relatively easy to find term by term. It might or might not be a polynomial or expressible in a closed form of some other given type. 

In the case of the central binomial coefficients, $F(x)$ is as you give in and $G(x)=1+2x+x^2.$
For the Fibonacci sequence the $F(x)$ is a rational function but the $G(x)$ doesn't appear at first glance to be anything nice.
A: Sure: choose any nonzero value for $a_0$, and write $F(x)=a_0+a_1x+a_2x^2+...$.
Expanding $F(x)^n$ gives you a LINEAR equation in $a_n$ as a function of the preceding ones, the coefficient of $a_n$ being $na_0^{n-1}$. In the special case of Fibonacci numbers, I do not know if $F(x)$ is "explicit", but formally it exists.
Added: choosing $a_0=1$ gives you
$$F(x)=1+x+x^2/2-x^3/3+x^4/8+x^5/15-25x^6/144+11x^7/70-209/5760x^8-319/2835
x^9 +...$$
and any other nonzero $a_0$ gives $a_0F(x/a_0)$.
Second addition: since some people seem interested in this expansion, two remarks. Call $a_n$ the coefficient of $x^n$. First, it must be easy to show
that the denominator of $a_n$ divides $n!$ (and even $(n-1)!$). Second, and much more interesting, is that the numerator of $a_n$ seems to be always smooth, more precisely its largest prime factor never exceeds something like $n^2$. This is much more surprising and may indeed indicate some kind of explicit expression.
Third addition: thanks to the answers of Fedor, Richard, and Ira, it is immediate to see that $F(x)$ is a solution of the differential equation
$y'-1=x/y$, giving the recurrence formula for the coefficients $c_n$ of $F$:
$$\sum_{0\le n\le N}(n+1)c_{n+1}c_{N-n}-c_N=\delta_{N,1}$$
Does this help ?
A: Lagrange–Bürmann formula claims that $$[w^{n-1}]H'(w)(\varphi(w))^n=n[z^n]H(g(z)),$$
where $g(z)$ and $f(w)=w/\varphi(w)$ are compositionally inverse power series without constant term (that is $g(f(w))=w$, $f(g(z))=z$), $H$ arbitrary power series. So if you choose $H$ equal to the antiderivative of $1/\varphi$, we get $$[w^{n-1}]\varphi^{n-1}=n[z^n]H(g(z))=[z^{n-1}](H(g(z))'=[z^{n-1}]\frac{g'(z)}{\varphi(g(z))}=\\ [z^{n-1}]\frac{g'(z)f(g(z))}{g(z)}=[z^{n-1}]\frac{zg'(z)}{g(z)}.$$
So, if you look for $\varphi$ such that
$$\sum_{n=1}^
\infty ([w^{n-1}](\varphi(w))^{n-1})z^{n-1}=u(z)$$
is a fixed function (satisfying $u(0)=1$), you should at first solve $zg'(z)/g(z)=u(z)$ that reads as $$(\log g)'=u(z)/z,\quad\log g(z)=\log z+\int_0^z \frac{u(t)-1}tdt+{\rm const},\\ g(z)=C z\exp\left(\int_0^z \frac{u(t)-1}tdt\right),$$
after that solve $$g(z)/\varphi(g(z))=z,\quad\text{i.e.}\,\, \varphi(s)=s/g^{-1}(s).$$
The choice of $C$ is your freedom.
You may try Fibonacci numbers, that is, $u(z)=1/(1-z-z^2)$. 
We get 
$$
g(z)=Cz(1-\alpha z)^{-\alpha/\sqrt{5}}(1-\beta z)^{\beta/\sqrt{5}},\quad
\alpha=\frac{1+\sqrt{5}}2,\,\beta=\frac{1-\sqrt{5}}2.
$$
The inverse map $g^{-1}$ is not explicit of course. You may probably look at it as a Christoffel–Schwarz type map (inverse of antiderivative of a product $\prod (z-z_i)^{c_i}$, where $c_i$ are all equal to $-1$ in our case).
A: Another way to state Fedor's answer is Exercise 5.56(a) of Enumerative Combinatorics, vol. 2. Namely, if $G(x)=a_1x+a_2x^2+\cdots$ is a power series (say over $\mathbb{C}$) with $a_1\neq 0$  and $n>0$, then
 $$ n[x^n]\log \frac{G^{\langle -1\rangle}(x)}{x} =
    [x^n] \left(\frac{x}{G(x)}\right)^n, $$
where $^{\langle -1\rangle}$ denotes compositional inverse. Letting both sides equal the Fibonacci number $F_n$ (using the indexing $F_1=F_2=1$) gives
  $$ F(x) =\frac{x}{\left( x\exp \sum_{n\geq 1}F_n\frac{x^n}{n}
       \right)^{\langle -1\rangle}}. $$
One can find a closed expression for $\sum F_n\frac{x^n}{n}$ by integrating $\sum_{n\geq 1} F_nx^{n-1}=1/(1-x-x^2)$, but there is no simple formula for the resulting compositional inverse.
