Re-reading your question, I think that I see what you are asking.

Per @Andrea Ferretti's comments, you have to be careful to distinguish between $\{e^{inx}\}$ and $span \ \{e^{inx}\}$. You certainly are interested the latter. Sorry if my comments were sloppy and confusing above.

So, I think that the it goes like this:

From some corollary of Stone-Weirstrauss you can show that $span \ \{e^{inx}\}$ is dense in $C(\mathbb{S}^1)$ with the supremum norm. Because we know that $C(\mathbb{S}^1)\hookrightarrow L^2([0,1])$ has its image a dense subset of $L^2([0,1])$ and we know that if $f_n \to f$ in the supremum topology on $C(\mathbb{S}^1)$, then the images also converge in $L^2([0,1])$.

Thus, by this reasoning, for $f\in L^2([0,1])$ we can find $f_n \in span \ \{e^{inx}\}$ such that $f_n = L^2([0,1])$. Lets write
$$
f_n = \sum_{k\in \mathbb{Z}} c_k^{(n)} e^{ikx}
$$
where all but finitely many of the $c_k^{(n)}$ are zero (this is because in the span of infinitely many objects we only take a finite number of them to add together)

Now, what I think you are asking is: what can we say about the coefficients $c_k^{(n)}$? The answer is that they converge to the $k$-th Fourier coefficient of $f$ as $n\to\infty$ because
$$
\hat f(k) = \langle f, e^{ikx} \rangle = \lim_{n\to\infty} \langle f_n ,e^{ikx}\rangle = \lim_{n\to\infty} c_k^{(n)}
$$

In fact if $c_k^{(n)}$ are arbitrary complex numbers, defining $f_n$ as above, we see that
$$
\Vert f - f_n \Vert_{L^2} = \sum_{k\in \mathbb{Z}} |\hat f(k) - c_k^{(n)}|^2
$$
assuming convergence. Thus, if $(c_k^{(n)})_k \to (\hat f(k))_k$ as $n\to\infty$ in $\ell^2(\mathbb{Z})$ then $f_n\to f$ in $L^2$, which is a pretty weak condition.

which$L^1$? All your functions are $2\pi$-periodic, so you cannot approximate anything that isn't. – Thierry Zell Aug 24 '10 at 13:27`$\{ e^{inx} \}$`

is trivially not dense in $L^1(0,1)$. What you probably mean is that their linear combinations (trigonometric polynomials) are. Still, this does not give a priori a development as a Fourier series, since the coefficients of the trigonometric polynomial approximation to a given function $f$ may not stabilize. – Andrea Ferretti Aug 24 '10 at 16:18