There is a unique invariant mean $m$ on WAP functions on any discrete group (see definitions below, theorem of ?). However, the proofs I found are fairly non-explicit on how to obtain this invariant mean.

By "explicit", I would like to know (preferably, for the free group on two generators, but other non-amenable groups are welcome), if there is a sequence of functions $f_n \in \ell^1\Gamma$ so that $f_n$ tends weak$^*$ to the mean $m$?

There is of course bound to be a big amount of non-explicitness due to the fact that, ultimately, elements of $(\ell^\infty \Gamma)^* \setminus \ell^1\Gamma$ require [a weak form of] the axiom of choice (the "Hahn-Banach axiom").

Here is an example of what I mean by explicit. When the group is amenable, uniqueness implies that this means coincides with any other invariant mean on $\ell^\infty\Gamma$. So it would suffice to pick a sequence $F_n$ of Folner sets, and let $f_n = \chi_{F_n} /|F_n|$ be the normalised characteristic functions of those sets.

As for the definitions...

$\Gamma$ acts on $\ell^\infty \Gamma$ by translation by the left: $(\gamma \cdot f)(x) = f(\gamma^{-1}x)$. (Likewise, it acts on the right.) A function $f$ is WAP if the weak closure of its orbit under this action is weakly compact. (WAP is a subspace of $\ell^\infty$ which contains $c_0$.)

An invariant mean on a subspace $X$ of $\ell^\infty \Gamma$ is a linear map $m\colon X \to \mathbb{R}$ so that $m(\chi_\Gamma) = 1$ (where $\chi_\Gamma$ is the constant function taking value $1$) $m(f) \geq 0$ if $f \geq 0$ and $m( \gamma \cdot f) = m(f)$ (likewise on the right). Linearity, positivity and $m(\chi_\Gamma) =1$ automatically implies that $m$ is bounded, i.e. $m \in (\ell^\infty\Gamma)^*$.

A [non-explicit] characterisation of the invariant mean on WAP functions goes as follows. Let $f$ be WAP. Then there is a unique constant function in the weak closure of its orbit under translations. The value that this constant function takes is the value of the mean.