(Essentially from Narutaka's comments)

Let $K = \cap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of $$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$

For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$ Indeed, if it's nonzero, we can find $(x_i)$ such that $$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$ then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.

Hence, if $f \in K$, then $$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$ has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.

The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.

(Essentially from Narutaka Ozawa's comments)

Let $K = \bigcap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of $$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$

For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$ Indeed, if it's nonzero, we can find $(x_i)$ such that $$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$ then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.

Hence, if $f \in K$, then $$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$ has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.

The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.

(Essentially from Narutaka's comments)

The trivial answer would be the closed linear span of $$\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G).$$

For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, this is the set of $f$ such that $$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$ Indeed, if it's nonzero, we can find $(x_i)$ such that $$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$ then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is nonzero at $f$. The converse follows from the fact that $$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$ has the same mean as $f$ for every invariant mean.

(Essentially from Narutaka's comments)

Let $K = \cap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of $$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$

For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$ Indeed, if it's nonzero, we can find $(x_i)$ such that $$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$ then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.

Hence, if $f \in K$, then $$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$ has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.

The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.