Using inclusion-exclusion, we can give the following explicit formula:
$$\mathrm{lcm}(2^1-1,2^2-1,\dots,2^n-1) = \prod_{k=1}^n (2^k-1)^{M(\lfloor n/k\rfloor)},$$ where $M(\cdot)$ is Mertens function.
Ignoring "$-1$" in the factors, the last expression can be approximated by $$2^{\sum_{k=1}^n kM(\lfloor n/k\rfloor)} = 2^{\Phi(n)} \approx 2^{\frac3{\pi^2}n^2},$$ giving the same estimate as in the other answers, where $\Phi(\cdot)$ is the totient summatory function. Perhaps, this approximation can be made rigorous, but I'll leave it here just as an idea.