Do we know any bound on $\operatorname{lcm}(2^1-1, 2^2-1,\dots,2^n-1)$?

Here my guess that seems to be confirmed by numerical computation.

Let $(u_n)_{n \geq 0}$ be a non-degenerate Lucas sequence of the first kind with a dominant positive real root $\alpha$. In other words, $u_0 = 0$, $u_1 = 1$, $u_{n+2} = a u_{n+1} + b u_n$ for each $n \geq 0$ for some relatively prime integers $a, b$; and the roots $\alpha, \beta$ of the characteristic polynomial $X^2 - aX - b$ are real and satisfy $\alpha > |\beta|$.

Can we find some asymptotic for $L_n := \text{lcm}\{u_k : k \leq n\}$ as $n \to \infty$?

Note that your case of $u_k = 2^k-1$ is given by $a = 3$, $b = -2$ (and hence $\alpha = 2$).

It holds the product formula $$u_n = \prod_{\substack{d \,\mid\, n \\ d > 1}} \Phi_d ,$$ where $$\Phi_d := \prod_{\zeta \text{ primitive $d$-th root of unity}} (\alpha - \zeta \beta)$$ is the homogeneous cyclotomic polynomial of $\alpha$ and $\beta$. Thus, clearly $L_n \mid \prod_{k \leq n} \Phi_k$. On the other hand, it is not true that the $\Phi_d$'s are pairwise coprime, however if $\gcd(\Phi_k, \Phi_h) > 1$ for some $k < h$, then $\gcd(\Phi_k, \Phi_h)$ is a prime number. Therefore, it seems quite possible that $L_n$ is not too smaller than $\prod_{k \leq n} \Phi_k$, so I guess $\log(L_n) \approx \sum_{k \leq n} \log(\Phi_k)$. Since $\alpha$ is dominant and $\Phi_k$ has degree $\varphi(k)$, it should be $\Phi_k \approx \alpha^{\varphi(k)}$. Finally, from the asymptotic formula $$\sum_{k \leq n} \varphi(k) = \frac{3}{\pi^2}n^2 + O(n)$$ my conjecture is that $$\log(L_n) \sim \frac{3\log\alpha}{\pi^2} \cdot n^2 ,$$ as $n \to \infty$.

P.S. My conjecture is true. The question was studied in much generality by S. Akiyama and F. Luca [1]. In particular, their Theorem 2 gives the guessed asymptotic for $L_n$, also with a good error term.

[1] S. Akiyama and F. Luca, On the least common multiple of Lucas subsequences, Acta Arith. 161 (2013), 327--349

The Szymiczek paper that I mentioned in comments some time ago is now available on the web. From Theorems 7 and 8 in the paper, we get $$\log{\rm lcm}(2^1-1,\dots,2^{[x]}-1)={3\log2\over\pi^2}x^2+O(x\log x)$$