One way to answer your question is to check if the following proof system is sound:
- Prover sends $R_1,\dots,R_n$
- Verifier sends challenge $c$
- Prover responds with $s_1,\dots,s_n$
- Verifier checks for each $i$ that $(R_i,c,s_i)$ is an accepting transcript of the Schnorr protocol
Note that this proves a different language, namely $L=\{(g_1,\dots,g_n)\,|\, \forall i\exists w_i:\,g_i=g^{w_i}\}$. Now, if the original protocol was special sound, which is the case for Schnorr's protocol, then this is also special sound. Given two accepting transcripts $(R_{1..n},c,s_{1..n})$ and $(R_{1..n},c',s'_{1..n})$ with $c\neq c'$, you can use the witness extractor of Schnorr's protocol for each $i$ on input $(R_i,c,c',s_i,s'_i)$ to recover every $w_i$.
So soundness is preserved, what about zero-knowledge? You can just invoke the special HVZK simulator for Schnorr's protocol $n$ times on input $c$ to get $n$ accepting transcripts of the form $(R_i,c,s_i)$. This should have the same distribution as a real transcript.
In short, the new protocol is also a $\Sigma$-protocol, i.e. it satisfies special soundness (it's a proof of knowledge), special HVZK and public randomness for the verifier. So applying the Fiat-Shamir transform to it should get you a non-interactive zero-knowledge proof of knowledge with provable security in the random oracle model.
