Brute Force on 3DES with Reduced Keyspace and Unknown IV
My first thought was that I could set the IV to the first 8 bytes of the CT [and] decode the rest[.]


This is exactly how CBC works.

For all blocks but the first, encryption is defined by $C_n=E_K(C_{n-1}\oplus P_n)$ and, therefore, decryption is achieved by $P_n=C_{n-1}\oplus D_K(C_n)$.

Since there is no previous ciphertext for the first plaintext block ($P_0$), we just set $C_{-1}=IV$. So, yes, considering the first ciphertext block the IV will allow you to decrypt all remaining ciphertext blocks.


  [T]hen substitute back to find the IV.


That part is not that easy. The first block of plaintext is $P_0=IV\oplus D_K(C_0)$. If both IV and plaintext are actual text, there might be hope, although 8 bytes of text probably won't be sufficient for the standard tools (e.g., frequency analysis).

If the IV has been generated pseudo-randomly, it effectively acts as a one-time pad. This means that, unless you can deduce the first 8 bytes of plaintext from the remaining plaintext anyway, there's nothing you can do.