One can still construct the space which we call $BG$ for example through the milnor space construction $BG=(G*G*G*...G)/G$ and it works for any topological group $G$. (There is a construction that gives a homotopy equivalent space to BG using simplicial methods that works more generally - for instance in constructing $B(BG)$).
We have the fundamental fact which you can find in Husemoller's Fiber Bundles page 57.(He does it in the setting of $G$ a topological group and $BG$ constructed using the milnor space construction).
For any $G$-principal fiber bundle $\xi \to X$ over a space $X$ with
a numerable cover, there is a map $X \xrightarrow{\phi} BG$ such that
the pullback of the fiber bundle $E_G \to BG$ along $\phi$ is $\xi$.
Since $E_G$ is contractible, the fiber of the homotopy fibration $X \to BG$ is $\xi$.
Apply this result to the principal $N$-bundle $N \hookrightarrow G \to G/N$ with $X=G/N$ to get a uniquely determined fibration up to homotopy.
Now repeat:
We have a $G$- principal fibration $G \hookrightarrow G/N \to BN$. We get a uniquely determined homotopy fibration $G/N \hookrightarrow BN \to BG$. Repeat again to get a fibration $BN \hookrightarrow BG \to B(G/N)$. One has the Serre spectral sequence of this fibration, $E_2=H^*(B(G/N),\mathscr{H}^*(BN))$ converging to $H^*(BG)$. This is the usual definition of the Lyndon-Hoschild spectral sequence.
If one decide that you are happy to use the Milnor space construction of $BG$ you can construct a fiber bundle $BN \to BG \to B(G/N)$ explicitly - see Alejandro Adem's book on the Group Cohomology and go the page on the Hoschild spectral sequence. He also works in the setting of $G$ a topological group.
So everything works as usual.
The only reason you might have had a doubt is that you might have learned, as in my answer Eilenberg-Maclane space $K(G\rtimes H, 1)$ for a semi-direct product., to get the fibration by saying that there is a map $BG \to BG/N$ whose fiber must be $BN$ by using the long exact sequence of homotopy groups. That derivation does not work obviously since $\pi_i(BG)\neq 0$ for $i\geq 1$.
