Well, the $n$th layer is a standard category.
We can even omit layers above $n$, or say they all contain only the identity cells.
What you wrote rather fits for (weak) $n$-groupoids (such as the fundamental groupoids of a topological space), where all cells are invertible.
In general, two $(n-1)$-cells $\alpha$ and $\beta$ are equivalent ($\alpha\simeq\beta$) in an $n$-category, if there is an invertible $n$-cell $\varphi:\alpha\to\beta$, i.e. there's also a $\psi:\beta\to\alpha$ such that $\psi\varphi=1_\alpha$ and $\varphi\psi=1_\beta$.
If $\alpha, \beta$ are $(n-2)$-cells, then we require $\simeq$ in the above equations in place of $=$.
And so on..
To see a specific example, consider the bicategory of rings and bimodules (with bimodules ${}_AM_B$ as arrows $A\to B$, tensor product as composition, and bimodule morphisms as 2-cells).
Here two bimodules are equivalent iff they are isomorphic, and two rings are equivalent iff there are bimodules between them, which are inverses to each other w.r.t tensor product (which is called they are Morita-equivalent).
Associativity is required on each layer up to (equivalence on that layer).
In the above example it's just $(M\otimes N)\otimes P\cong M\otimes (N\otimes P) $. Note that these entities are indeed not identical, for the same reason as $(A\times B)\times C\ne A\times(B\times C)$ for sets, but there's a natural isomorphism between them.
However, this condition for associativity in itself proved to be not sufficient, and further coherence conditions had to be posed, see e.g. the definition of a bicategory.
In weak higher categories these coherence issues are highly nontrivial.
The definition on Wikipedia is correct, and it's different from being Hausdorff.
Let's look at an example, the cofinite topology on $\Bbb N$, where the open sets are the subsets of $\Bbb N$ whose complement is finite. Given $m,n\in\Bbb N$ we can find a neighbourhood of $m$ not containing $n$, such as $\Bbb N\setminus\{n\}$, and a neighbourhood of $n$ not containing $m$, such as $\Bbb N\setminus\{m\}$. This space however is not Hausdorff since every two nonempty open sets have nonempty intersection, this example should convince you that the two notions are different.
The definition on Wikipedia for $T_1$ is equivalent to that of Munkres, I suggest you convince yourself of the equivalence as an exercise, but it has been asked on this website before if you want to look it up.
Best Answer
This answer may not be in the spirit of the question, but I think it's worth noting.
The usual definition of Hausdorff uses the notions "point" and "open sets". But both of these notions can be expressed in the language of categories: a point of $X$ is an arrow from the one-point space $1\to X$, and an open set of $X$ is an arrow from $X$ to the Sierpiński space $X\to S$. So you can translate the usual definition to the language of categories.
Recall that $S=\{0,1\}$ with open sets $\varnothing$, $\{1\}$, and $S$. Overloading notation, write $1$ for the arrow $1\to S$ mapping the unique point of $1$ to $1$.
$X$ is Hausdorff if and only if: for every pair of arrows $p_1\neq p_2\colon 1 \to X$, there is a pair of arrows $u_1,u_2\colon X\to S$ such that (1) $u_1\circ p_1 = u_2\circ p_2 = 1$, and (2) there is no arrow $q\colon 1\to X$ such that $u_1\circ q=u_2\circ q=1$.