I would like to calculate the $\mathbb{Z}$-Homology of $K(\mathbb{Z}/n,1)$ via the Fibration $$K(\mathbb{Z},1)\hookrightarrow K(\mathbb{Z},1)\twoheadrightarrow K(\mathbb{Z}/n,1)$$ and the Serre-Spectral-Sequence.
If we knew, that the local coeffitient system in $H_p(K(\mathbb{Z}/n,1);H_q(K(\mathbb{Z},1)))$ was trivial, then we could easily apply the Serre-Spectral sequence and derive $$H_p(K(\mathbb{Z}/n,1);\mathbb{Z})=\begin{cases}\mathbb{Z}&p=0\\ \mathbb{Z}/n &p=2k+1 \\ 0& else\end{cases}$$ The Problem here is obviously the non-trivial fundamental group of $K(\mathbb{Z}/n,1)$. This Problem is easily resolved for odd $n$ as there is no non-trivial morphism $$\mathbb{Z}/n\rightarrow Aut(\mathbb{Z})=\lbrace\pm 1\rbrace$$ for odd $n$. For even $n$ there is a non-trivial morphism, so we can not proceed using only formal arguments. Using Künneth one can assume $n=2^k$ if one wishes, but this also does not solve the real problem.
Is there some way to deduce the triviality of the local coeffitients (and are they trivial at all)?
Let's use the definition of $BG$ from Hatcher, built out of simplicies $[g_0,\ldots, g_n]$ whose vertices are in $G$, glued together along faces $[g_0,\ldots, \hat g_i,\ldots, g_n]$, and taken modulo the group action $g[g_0,\ldots, g_n] = [gg_0,\ldots, gg_n]$.
In order to compute the monodromy action on the fibers, we should find explicit paths in $BG$ representing the classes of $\pi_1(G)$. For $g\in G$, there is the linear path $\gamma: I \to EG$ along the 1-simplex $[1,g]$, whose projection in $BG$ is the loop in $\pi_1BG$.
Now, if we have a surjection of groups $H \to G$ and we wish to understand the monodromy of the fibration $BH \to BG$, then we need to lift the loop $[1,g] \in \pi_1(BG)$ to a path in $BH$. But it is easy to make this lift, simply take $\tilde g \in H$ and use $[1,\tilde g]$ which is a loop in $BH$. Since it's a loop, the induced action on the fiber is trivial.
The action on the homology groups is induced by the monodromy action, so the corresponding local coefficient system is trivial too.