Encrypted counter as IV in CBC mode encryption using same key
Just to be sure we're on the same page, I interpret your question as defining encryption of a string $P_1 P_2 \cdots P_\ell$ with a counter $\mathsf{ctr}$, key $K$, and an $n$-bit blockcipher $E$ as follows:
$$ \mathcal{E}_K(\mathsf{ctr}, P_1P_2\cdots P_\ell) = C_0 C_1 C_2 \cdots C_\ell$$
where $C_0 = E_K(\mathsf{ctr})$, $C_{i+1} = E_K(C_i \oplus P_{i+1})$, and each $P_i$ is an $n$-bit string. (So I'm assuming plaintexts have already been padded as necessary.) I'm assuming the values that will be used for $\mathsf{ctr}$ are determined ahead of time. If this is incorrect, attacks become possible.

The scheme above is secure in the sense of real-or-random security under chosen-plaintext attacks (this is a somewhat stronger condition than semantic security). That is, an adversary cannot distinguish between $\mathcal{E}_K$ and an oracle that given an nonce and an $nm$-bit plaintext returns $(m+1)n$ random bits. This is true as long as $E$ is secure (in the sense of being a pseudo-random permutation), $K$ is random, and the total length of all the ciphertexts encrypted under a given key is much less than $2^{n/2}$ bits (the birthday bound).

Proof sketech. A standard game-hopping security proof might proceed like this:


Change the scheme to use a random permutation $\pi$ in place of $E_K$ (this is reasonable if $E$ is secure and $K$ is random).
Change $\pi$ to a random function $\rho$ (this is reasonable as long as long as the number of queries to $\pi$ is much less than $2^{n/2}$; see the PRP-PRF switching lemma).


As long as no input to $\rho$ is ever repeated, all its outputs are uniformly random and independent. Therefore using the encryption scheme over $\rho$ is indistinguishable from an oracle that outputs random strings, as long as this condition holds. So we need to bound the probability that an input to $\rho$ is repeated.

From here the proof proceeds like the standard CBC security proof, with the caveat that instead of looking for collisions just among the $\rho$ inputs we've already seen, we also need to worry about one of the inputs being equal to a counter value we'll need in the future. (This is where the assumption that counter values are pre-determined comes into play.) Since barring a repeated input, each $C_i$ and therefore each $C_i \oplus P_{i+1}$ is a uniform random value independent of previous queries, the probability that a given $C_i \oplus P_{i+1}$ is equal to a counter value is at most $q / 2^n$, where $q$ is the number of queries the adversary will make (and hence the number of counter values we require).

Using a union bound, the probability that such a collision will ever occur is at most $q^2 M /2^n$, where $Mn$ is the length of the longest plaintext. This probability is negligible as long as $(qM) \ll 2^{n/2}$. And note that $qMn$ is an upper bound for the total number of bits encrypted.

So. The encryption scheme $\mathcal{E}$ is not significantly less secure than CBC with a random IV. The larger question of whether the protocol as a whole is secure as a much more difficult question, and one that I can't answer without knowing more about it (and probably couldn't answer even if I did). But you are authenticating the counter, right? :)