Why sample size is not a part of sufficient statistic? Following simple example from Wikipedia's definition of sufficient statistic with Bernoulli distribution with parameter $\theta$, where sufficient statistic is a sum of successes
$$T(X_n)=\sum_{i=1}^n x_i, x_i \in \{0,1\}$$
As MLE is $\frac{1}{n}\sum_{i=1}^n x_i$ we can't just use the sufficient $T(x)$, but we also need the sample size, which is a property of our sample, hence $T(x)$ doesn't contain all information about the sample. Is there an unspoken rule, that the sample size is not considered for the sufficient statistics?
 A: It is typical practice that the sample size is considered (implicitly) to be a known constant unless we specify the contrary in the analysis.  This practice saves time by alleviating the need to specify that the sample size is known, which is true in the vast majority of statistical applications.  You can of course proceed on the basis that $n$ is also an unknown parameter in the model.  In this latter case your log-likelihood function would be:
$$\ell_{\mathbf{x}_n}(n,\theta) = \log {n \choose T(\mathbf{x}_n)} + T(\mathbf{x}_n) \log(\theta) + (n-T(\mathbf{x}_n)) \log(1-\theta),$$
and the minimal sufficient statistic is indeed $(n,T(\mathbf{x}_n))$ (so the statistic $T(\mathbf{x}_n)$ is not sufficient in this case).
(Note: I do not agree with the comment by Xi'an asserting that $n$ must be the outcome of a random variable to be included as part of the sufficient statistic; the concept of sufficiency is a classical concept, and in that domain the notion of an "unknown constant" is perfectly valid.  There is no need to create a Bayesian model that specifies a distribution for $n$ in order for it to be part of the sufficient statistic.)
