Heuristically, we should expect only finitely many such numbers. If $C(x)$ is the number of highly composite numbers which are less than or equal to $x$, then Erdos and Nicholas showed that there are positive constants $a$ $b$, $K_1$ and $K_2$, such such that $1 < a< b$ and $K_1 (\log x)^a < C(x) < K_2(\log x)^b $ . So the $n$th highly composite number grows at least as fast as $ e^{n^T}$ for some constant $T>1$. Then by the prime number theorem, the chance that we have two twin primes on either side should be roughly $$\left(\frac{1}{\log e^{n^T}}\right)^2 = \frac{1}{n^{2T}}.$$
But $\sum_{n=1}^{\infty} \frac{1}{n^{2T}}$ converges so we expect there to be only finitely many $n$ that work.
This heuristic may not be completely persuasive because highly composite numbers have a lot of prime factors, so numbers 1 away from them should have a slightly higher chance of being prime than one would randomly expect. However, this should alter things only by a small factor, likely not enough to change the overall behavior.