Implications of reversing CBC encryption
If I understand you correctly, you want to use the decryption process of CBC to encrypt, i.e., 



This would not be CPA secure. The problem is, as you already noted, that the IV has no influence on any block but the first one.

I.e. a CPA attacker works as follows:
Choose two random messages of length 2 blocks $(m_0^0,m_0^1)$ and $(m_1^0,m_1^1)$ and output them as the challenge messages. Receive back the challenge ciphertext $(c^0,c^1)$. Now, ask $(m_0^0,m_0^1)$ to the encryption oracle and receive the encryption $(c_0^0,c_0^1)$. If $c^1 = c_0^1$ then output $0$, otherwise output $1$.