How reassuring is CRC-then-XTS as authenticated encryption?
No, this does not make a deterministic authenticated cipher, unless you're using a secret CRC as a MAC.

How do you break it?  First, find a pair of messages $m = m_1 \mathbin\| m_2$ and $m' = m'_1 \mathbin\| m'_2$ so that $m_i \ne m'_i$ and $\operatorname{CRC}(m) = \operatorname{CRC}(m')$.  (Finding collisions in CRCs is not hard.)  Then:


Query the oracle for $m_1 \mathbin\| m_2$ to find the ciphertext $c_1 \mathbin\| c_2 \mathbin\| c_3$; assume the CRC is in the plaintext of block $c_3$.
Query the oracle for $m'_1 \mathbin\| m_2$ to find the ciphertext $c'_1 \mathbin\| c_2 \mathbin\| c_{\mathit{garbage}}$.
Query the oracle for $m_1 \mathbin\| m'_2$ to find the ciphertext $c_1 \mathbin\| c'_2 \mathbin\| c_{\mathit{garbage}}$.


Finally, yield $c'_1 \mathbin\| c'_2 \mathbin\| c_3$ as a forgery, for the message $m'_1 \mathbin\| m'_2$ not previously sent to the oracle.