Question: How can I simplify $E( I\_1\cdot I\_2\cdot \dots \cdot I\_n )$, which is a mean of the product of $n$ ***dependent*** indicator random variables. Can I write down something like $E( I\_1\cdot I\_2\cdot \dots\cdot I\_n )=E( I\_1)\cdot E(I\_2)\cdot \dots \cdot E(I\_n)+\operatorname{Cov}$ in which $\operatorname{Cov}$ is a "covariance" term(s)?

I know there is no such answer for $E\left( X^n\right)$, in which $X $are random variables. But my case is about n different indicator random variables, so I am having a glimmer of hope to simplify the math.

Answer: 
The mean (i.e., expected value) of the product of n dependent indicator random variables equals the probability of the event constituting the intersection (i.e., simultaneous occurrence) of the events underlying all the indicator random variables. If you know that probability, you have the answer, otherwise you don't. 

Without knowing anything about the dependency among the $I\_i$'s, this probability can be bounded as being in $[0,\text{min}\{ E(I\_1),...,E(I\_n) \} ]$, where of course $E(I\_i) = P(\text{event for which } I\_i \text{ is the indicator})$.
Let $A=\prod\_i I\_i$, then $A=1$ or $A=0$, and $A=1\iff I\_i=1\forall i$.

Now, $E[I\_1\cdot I\_2\cdots I\_n]=E[A]=1\cdot P(A=1) + 0\cdot P(A=0)=P(A=1)$

and $P(A=1)=P(I\_1=I\_2=...=I\_n=1)$, so you only care about the even in which all of them are equal to one together. It's hard to say more without knowing anything about the distribution.