Considering a key of 128bit and a message of 128bit, is it possible to recover the key (e.g. if the message is all zeros)?
Yes, but only by brute-force. That is when given a $(m,\tau)$ message-tag pair for a deterministic MAC (such as CMAC), you can always brute-force the key $k$ to get $\tau=\operatorname{MAC}_k(m)$. However with keyspace $\mathcal K$, this will take $|\mathcal K|/2$ MAC calls on average, which is $2^{127}$ evaluations which are usually assumed to be infeasible.
Which is the suggested minimum message length compared to the key length?
There is no minimum message length. In fact, quite the opposite is true, MAC security degrades with message length. However, short messages may be more susceptible to replay attacks, but it's the task of the higher-level protocol to deal with them.
Do you suggest some requirements in order to avoid forgery attacks?
The security bound for CMAC as a MAC is
$$\mathbf{Adv}^{\textsf{MAC}}_{\operatorname{CMAC}(\pi)}(\mathcal A;q,ln)\leq \frac1{2^n}+\mathbf{Adv}^{\textsf{PRP}}_\pi(\mathcal B;\sigma)+\frac{2.5(\sigma+1)^2+1.5q^2+2q^2l^2}{2^n}$$
for an adversary, $\mathcal A$ making $q$ signature queries each of length at most $ln$ bits - with $n$ being the block size of the underlying block cipher $\pi$ - resulting in a total of at most $\sigma$ different evaluations of the block cipher $\pi$.
$\mathbf{Adv}^{\textsf{MAC}}_{\operatorname{CMAC}(\pi)}(\mathcal A;q,ln)$ is the chance that an adversary $\mathcal A$ can forge a tag for a message where they did not query the signing oracle beforehand and using at most $q$ queries each of length at most $ln$ bits.
$\mathbf{Adv}^{\textsf{PRP}}_\pi(\mathcal B;\sigma)$ is the probability that the block cipher $\pi$ can be distinguished from a random permutation using $\sigma$ encryption queries. This bound was proven (written slightly differently) in "Formal Security Proof of CMAC and Its Variants" by Baritel-Ruet, Dupressoir, Fouque, and Grégoire in 2018 (PDF).
