It took some effort, but I found a closed expression.
First, we use the close form and roots of the Laguerre polynomials to express the integral as:
\begin{equation}
f_n=\frac{1}{2}\sum_{q=0}^n\sum_{\lambda\in\textrm{Roots}(L_n)}\left(\begin{array}&n\\q\end{array}\right)\frac{(-2)^q}{q!}\int_0^\infty e^{-r^2} r^{2q+1}\log\left[\frac{(2r^2-\lambda)^2}{\lambda^2}\right]\,dr
\end{equation}
To carry out this integral, we integrate by parts and make use of the change of variable $x=r^2-\frac{\lambda}{2}$:
\begin{multline}
\int_0^\infty e^{-r^2} r^{2q+1}\log\left[\frac{(2r^2-\lambda)^2}{\lambda^2}\right]\,dr=4\int_0^\infty \frac{r\, \Gamma\left(q+1,r^2\right)}{2r^2-\lambda}\, dr\\
=2q!\sum_{l=0}^q\frac{1}{l!}\int_0^\infty\frac{r^{2l+1}e^{-r^2}}{r^2-\frac{\lambda}{2}}\, dr=q!\sum_{l=0}^q\frac{e^{-\frac{\lambda}{2}}}{l!}\int_{-\frac{\lambda}{2}}^\infty \frac{e^{-x}}{x} \left(x+\frac{\lambda}{2}\right)^ldx
\end{multline}
Using the binomial formula, we have:
\begin{multline}
\int_{-\frac{\lambda}{2}}^\infty \frac{e^{-x}}{x} \left(x+\frac{\lambda}{2}\right)^ldx=\sum_{k=0}^l \int_{-\frac{\lambda}{2}}^\infty e^{-x}\left(\begin{array}&l\\ k\end{array}\right)\left(\frac{\lambda}{2}\right)^{l-k}x^{k-1}\, dx\\
=-\left(\frac{\lambda}{2}\right)^l Ei\left(\frac{\lambda}{2}\right)+\sum_{k=1}^l\left(\begin{array}&l\\ k\end{array}\right)\left(\frac{\lambda}{2}\right)^{l-k}\Gamma\left(k,-\frac{\lambda}{2}\right)\\
=-\left(\frac{\lambda}{2}\right)^l Ei\left(\frac{\lambda}{2}\right)+
e^{\frac{\lambda}{2}}\sum_{k=1}^l\sum_{s=0}^{k-1}\frac{l!\,(-1)^s}{s!\,(l-k)!\,k}\left(\frac{\lambda}{2}\right)^{l-k+s}
\end{multline}
Putting all the pieces together, we finally obtain:
\begin{multline}
f_n=\sum_{q=0}^n\sum_{\lambda\in\textrm{Roots}(L_n)}\sum_{l=0}^q\left(\begin{array}&n\\q\end{array}\right)\left[\frac{(-1)^{q+1}\lambda^l\, 2^{q-l-1}}{l!}e^{-\frac{\lambda}{2}}Ei\left(\frac{\lambda}{2}\right)+\sum_{k=1}^l\sum_{s=0}^{k-1}\frac{(-1)^{s+q}\,\lambda^{l-k+s}\,2^{q-l-s+k-1}}{s!\,(l-k)!\,k}\right]
\end{multline}
