I'm by no means an expert, but I would think you can let $\operatorname{disc}(P_n(t))^{1/d_n}$
approach infinity by letting the leading coefficient grow fast enough.

For example, just to be cute, let $n > 2$, let $F_n$ the $n$-th Fibonacci number and let $G_n = \prod_{i < j \leq n}(F_i-F_j)$. Now, pick another sequences $s_n = (nG_n)^{n^2}$ and define our polynomials $P_n(t) = s_n(t - F_1)(t - F_2)\cdots(t - F_n)$ for all $n > 2$ (so $d_n = n$ for us). Then

$\operatorname{disc}(P_n(t))^{1/d_n} = \left[s_n^{2n-2}\prod_{i < j \leq n} (F_k - F_j)^2\right]^{1/n} = \left[(nG_n)^{n^2(2n-2)}\prod_{i < j \leq n} (F_k - F_j)^2\right]^{1/n}$

But this is just

$n^{n(2n-2)}G_n^{2n-2}\prod_{i < j \leq n} (F_k - F_j)^{2}$

which tends to infinity quite quickly.