There is a missing exponent in your trivial bound. Letting $u_n = \sum_{k=1}^n \sqrt{p_k}$, we have
$$B_0(n) \equiv \max_i \; \left| [x^i] S_n \right|$$
$$\le B_1(n) \equiv \max_i \; [x^i] (x+u_n)^{2^n} = \max_i {2^n \choose i} u_n^i$$
$$\le B_2(n) \equiv { 2^n \choose 2^{n-1} } u_n^{2^n}$$
$$\le B_3(n) \equiv 2^{2^n} (n \sqrt{p_n})^{2^n}.$$
Of course, in the last step we can get a better bound by estimating $u_n$ less crudely. It should also be possible to find an analytic bound for $B_1(n)$ that is less crude than $B_2(n)$ (regardless, $B_1(n)$ is easy to evaluate numerically).
For reference, I have computed numerical approximations of the actual value $B_0(n)$ up to $n = 20$ (correct up to rounding in the last digit):
0: 1
1: 2
2: 10
3: 960
4: 13950764
5: 2.55690851718529e+17
6: 1.77108072043063e+39
7: 8.57834471403602e+86
8: 4.69693103314689e+187
9: 3.24515842436673e+401
10: 8.31078370973184e+853
11: 4.18601441612784e+1793
12: 1.37441368638541e+3755
13: 7.32398012717744e+7815
14: 4.7364530607185e+16172
15: 8.41442697691365e+33355
16: 6.50154879984207e+68684
17: 2.98829955473397e+141188
18: 7.8161464597922e+289271
19: 2.1050522533847e+591950
20: 2.92232330678161e+1209132
Here is a comparison of the bounds:
It appears that $B_3$, $B_2$ and $B_1$ asymptotically overestimate the number of bits in $B_0$ less than by a respective factor 2.20, 2.02 and 1.76.
As you say, it should possible to do better by taking into account the cancellation that occurs. Perhaps by expressing the coefficients in terms of elementary symmetric polynomials? A lower bound would also be interesting.
