MuSig/Schnorr key aggregation: How get the combined private key Given all parties' private keys in MuSig, is it possible to get a combined private key corresponding to the combined public key? I know it's possible to sign a message, but how to get the combined private key?
Yes, it's possible if you eventually know all the parties' secret keys! However, note, that in most realistic application scenarios, where there is at least one honest party whose secret key is not known, it is computationally infeasible to recover the aggregated secret key, under the  Discrete–Logarithm assumption,  given the secret keys of the corrupted parties. In the following let me use the notations introduced in the MuSig paper by Maxwell et al.

Let $L$ be a multiset (the same public key can appear multiple times in $L$) of signers' public keys: $$L=\{X_1=g^{x_1},\dots,X_n=g^{x_n} \}. $$ The message to be signed is denoted as $m$, $H$ denotes a hash function and $R=\prod_{i=1}^n R_i=\prod_{i=1}^n g^{r_i},$ where $r_i$ is a share generated by $i$th participant. Challenges are computed as: $c_i=H_{agg}(L,X_i)\cdot H_{sig}(\tilde{X},R,m),$ where we define $a_i= H_{agg}(L,X_i)$ and the aggregated public key is defined as $\tilde{X}=\prod_{i=1}^n X_i^{a_i}.$ 

Since $\tilde{X}=\prod_{i=1}^n X_i^{a_i}=g^{\sum_{i=1}^n a_i x_i}$ Therefore it is easy to see that the aggregated secret key is $\sum_{i=1}^n a_i x_i.$ The verification equation for an $(R,s)$ aggregated signature on message $m$ is: $$g^s=R\tilde{X}^c, $$ where $c=H_{sig}(\tilde{X},R,m)$.
Although note that the aggregated secret key has no role in verifying aggregated signatures and does not play any role in the MuSig scheme.