How to calculate the key size of my cryptographic system
To compute bit length equivalent of your key, you need to compute $\log_2(\Omega)$ where $\Omega$ is the space of possible keys. Now, I have myself been computing the equivalent key sizes of many non-modern cryptosystems so it proved a really interesting exercise.

The formula I'm left with is

$$
\log_2(1147^N \cdot 94^N \cdot 2^N \cdot 2)
$$
where $N$ is the number of rotors.

First, we select the rotors. For $N$ rotors, there are $1147^N$ possible choices. Then, we select their initial state (which were historically distributed by a codebook, different for each day). For each rotor we have $94$ possibilities, therefore for the whole setup there are $94^N$ possibilities. Then, each rotor can have two directions, hence $2^N$. The flow direction, I assume, is global, therefore we finally multiply all of this by $2$.

So, for the calculation, we have

$$
\log_2(1147^N \cdot 94^N \cdot 2^N \cdot 2)\\
= \log_2(1147^N \cdot 94^N \cdot 2^N)  +  \log_2(2)\\
= N \cdot \log_2(1147 \cdot 94 \cdot 2) + \log_2(2)\\
\approx N \cdot 17.718 + 1
$$

You will therefore have over 128-bit keyspace with 8 rotors and you will need 15 of them for 256-bit keyspace.

As previously noted by Adrian K Luis Casillas [edit], it doesn't mean this cryptosystem has this much "bits of security". This notion is much less strict and takes cryptanalysis into account, e.g. a cryptanalyst can find the attack which, after collecting sufficient material, will reduce the effective key space by either a couple of bits or even a large factor, halving it or worse.

Note that if you have any objections for this analysis I will happily discuss this topic further. It has been fun.