It's easier to understand what's going on here if we talk in more generality, so as to avoid getting bamboozled by all the trig functions (plus, it's useful to know about these ideas for other bases).
The norm (or at least, the one it means) is defined as
$$ \lVert f \rVert_2 = \sqrt{\int_{-L}^L f(x)^2 \, dx}, $$
and there is an associated inner product $\langle f,g \rangle = \int_{-L}^L f(x) g(x) \, dx$, so $\langle f,f \rangle = \lVert f \rVert_2^2 $.
Suppose we want to have an expansion of $f$ in terms of an orthogonal basis of functions $ (e_n)_{n=0}^{\infty}$ (orthogonal means $\langle e_m,e_n \rangle=0$ if $n \neq m$),
$$f(x) = \sum_{n=0}^{\infty} A_n e_n(x)$$
(this is what a Fourier series is, if we choose $e_n$ to include $\frac{1}{2}$, $\cos{(k\pi x/L)}$ and $\sin{(k\pi x/L)}$). We find $A_n$ using the orthogonality:
$$ \langle f, e_n \rangle = \sum_{m=0}^{\infty} A_m\langle e_m,e_n \rangle = A_n \langle e_n , e_n \rangle = A_n \lVert e_n \rVert_2^2 $$
Thus
$$f = \sum_{n=0}^{\infty} \frac{\langle f, e_n \rangle}{\lVert e_n \rVert_2^2} e_n. $$
Parseval's identity gives the norm of $f$ in terms of the coefficients $A_n$:
\begin{align}
\lVert f \rVert_2^2 &= \langle f , f \rangle \\
&= \left\langle \sum_{n=0}^{\infty} A_n e_n , \sum_{m=0}^{\infty} A_m e_m \right\rangle \\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} A_n A_m \langle e_n , e_m \rangle \\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} A_n^2 \lVert e_n \rVert^2.
\end{align}
To get back to your problem, the important thing is that this includes the norm of the basis functions $e_n$. Therefore if the basis functions are not normalised to have $\lVert e_n \rVert^2=1$, you'll get an extra factor. Going back to Fourier series explicitly, the basis functions are not normalised in the usual formulation:
$$ \int_{-L}^L \cos^2{\left( \frac{n\pi x}{L} \right)} \, dx = \int_{-L}^L \sin^2{\left( \frac{n\pi x}{L} \right)} \, dx = L, \\
\int_{-L}^L \frac{1}{2} \, dx = L. $$
Hence the extra factor of $L$.
