Are there any symmetric cryptosystems based on computational complexity assumptions? Are there any symmetric cryptosystems which are provably secure in the sense that there exists a reduction from their security to the hardness of some underlying hard problem such as integer factorisation? 

If not, why not?
Finding uniform random $x$ given $x^3 \bmod{pq}$ for uniform random 1024-bit primes $p$ and $q$ is conjectured to be hard because smart, motivated cryptanalysts have spent decades trying to do so and have left only a track record of failure.*
Finding uniform random $k$ given $\operatorname{AES256}_k(92187681)$ is conjectured to be hard because smart, motivated cryptanalysts have spent decades trying to do so and have left only a track record of failure.


That said, the best estimates for the cost of (1) are much cheaper than the best estimates for (2), and the computation of $x^3 \bmod{pq}$ is much costlier than the computation of $\operatorname{AES256}_k(92187681)$.  In other words, RSA-2048 is much more expensive for less security than AES-256.

You might be tempted to say that the RSA problem is a more fundamental problem in number theory, and as such is the only one that's really a ‘hard problem’.  But it is precisely because RSA is embedded in a rich mathematical theory—as is needed for separate public key and private key operations!—that it is more vulnerable to attacks.  In reality, AES is a much harder problem than RSA!

There are many symmetric cryptosystems that use AES, and for which there is a theorem that breaking them can't be much easier than breaking AES, such as AES-GCM.  Similarly, there are many public-key cryptosystems that use the RSA trapdoor permutation, and for which there is a theorem that breaking them can't be much easier than inverting the RSA trapdoor permutation, like RSA-PSS and RSA-KEM.

The term ‘provable security’ means nothing more than there is a theorem.  These cryptosystems—AES-GCM, RSA-PSS, and RSA-KEM alike—all have ‘provable security’ because there is a theorem, not because of any mathematical theory around AES or RSA.  So does a 1-bit universal hashing authenticator have provable security, even though the amount of security it provides is so small an attacker will win with the probability of a fair coin toss coming up heads.



* Incidentally, while the RSA problem can't be harder than factorization, we don't have a proof that it can't be easier.  There is some weak evidence—a reduction in the generic ring model—but there's no theorem that if factoring is hard then the RSA problem is hard.  Hence not even the RSA problem has ‘provable security’ relative to factoring.