The microcanonical ensemble is the (maximum entropy) probability distribution for a given specified total energy. What you've calculated is actually the maximum entropy distribution with no constraints on the energy, which is the same as the canonical distribution at infinite temperature ($\beta = 0$).
To correctly calculate the microcanonical entropy, first observe that in this system, the total energy $E=\epsilon n_1$, or $n_1 = E/\epsilon$, as long as $E$ is an integer multiple of $\epsilon$.
Since the total energy completely determines $n_1$, we can calculate the number of states in the microcanonical ensemble just as you did, but without the summation over $n_1$:
$$\Omega(E) = \frac{N!}{(N-E/\epsilon)!(E/\epsilon)!}$$
This is only defined when $E$ is an integer multiple of $\epsilon$, but in an appropriate limit of large $N$ and small $\epsilon$ you should be able to use Stirling's approximation to write $S(E) = \log \Omega(E)$ as if it were a continuous function of $E$.
Once this is done, the temperature enters through the usual thermodynamic definition,
$$
\frac{1}{T} = \frac{\partial S}{\partial E},
$$
which ought to be fairly straightforward to calculate.
I hope this helps, and sorry for not working through all the details myself.
This answer is just summarizing the material on wikipedia.
Suppose there are $M$ energy levels, $E_1, E_2, \cdots, E_M$, and that we have $N$ distinguishable particles. For now, suppose that the energy levels are not degenerate. In other words, suppose there is one state associated with each energy level.
Then, we want to count the number of ways we can have $N_1$ particles in energy level 1, $N_2$ particles in energy level 2, etc.
There are
\begin{equation}
W = {N \choose N_1} = \frac{N!}{(N-N_1)!N!}
\end{equation}
ways $W$ to select $N_1$ particles to place in the first energy level. Then there are
\begin{equation}
W = {N-N_1 \choose N_2} = \frac{(N-N_1)!}{(N-N_1-N_2)! N_2!}
\end{equation}
ways to choose $N_2$ particles to go into the second energy level, from the remaining $N-N_1$ particles.
Multiplying these results together, we find there are
\begin{equation}
W= {N \choose N_1} {N-N_1 \choose N_2} = \frac{N!}{(N-N_1)!N!} \frac{(N-N_1)!}{(N-N_1-N_2)! N_2!} = \frac{N!}{N_1! N_2! (N-N_1-N_2)!}
\end{equation}
ways to have $N_1$ particles in the first energy level and $N_2$ particles in the second energy level.
Here, I will share a screenshot of the wikipedia article I linked above, which nicely shows the cancellation that occurs when computing the number of ways $W$ of having a given set of occupation numbers
The final step here is to realize that the last factor in the denominator in the second line is simply 1, since $N$ is equal to the sum of the occupation numbers in each state, so $N-\sum_{i=1}^M N_i = 0$ and then we use $0!=1$.
This leads to the result
\begin{equation}
W = N! \prod_{i=1}^M \frac{1}{N_i!}
\end{equation}
when ignoring degeneracies.
Now we include the possibility that energy level $i$ has a degeneracy of order $g_i$. This means that there are $g_i$ states that have the same energy $E_i$.
To account for this hiccup, consider what happens when we place the first of $N_i$ particles into the energy level $E_i$. There are $g_i$ choices for which state we place this particle into. Then we place the second particle, and again there are $g_i$ choices, so there are $g_i^2$ choices for placing particles 1 and 2. Carrying out this procedure for all $N_i$ particles, we find that there are $g_i^{N_i}$ ways of placing $N_i$ particles into this energy level.
Applying this correction to the previous result for $W$, we get the final result
\begin{equation}
W = N! \prod_{i=1}^M \frac{g_i^{N_i}}{N_i!}
\end{equation}
Loosely speaking, the $1/N_i!$ factor is there to count the number of ways to place $N_i$ particles into an energy level $E_i$, and the $g_i^{N_i}$ factor is there to account for the number of ways to assign $N_i$ particles to the $g_i$ states within that energy level.
Best Answer
The density of states is defined that way, such that in a continuous distribution (which is valid here because the energy is large), the number of states between $E_1$ and $E_2$ is
$$N = \int_{E_1}^{E_2}dE \ \Omega(E)$$
where $\Omega(E)$ is the number of microstates for energy $E$. In the case of the infinitesimal range $\delta E$, the function varies negligibly over the range, so the (differential) number of states (leaving out mathematical rigor) is simply the DOS multiplied by the range $\delta E$. That's just mathematics. The same goes for charge density in electrostatics, fluid density, etc.