The answer actually depends more on the space. And upper bounds on the modal probability are not very interesting unless you put conditions on the random variables not to be too close to deterministic; if you put such conditions then you get answers of more or less the same shape as the lower bounds on the modal probability, so let's look at that.

Let's stick to random variables $X_i$ which take values $0$ and $x_i$, since this captures the problem fairly well. The simplest case is $x_i$ doesn't depend on $i$, i.e. you are working in a $1$-dimensional space. Then it is fairly easy to see that the 'best case' is $p_i$ identically $1/2$, meaning this minimises the modal probability: compute the moment generating function and use Jensen's inequality.

If on the other hand the $x_i$ are linearly independent (or just all sums are distinct, even if the dimension is $1$!), then the modal probability is just the product of $\max(p_i,1-p_i)$, again minimised when all $p_i$ are $1/2$, but this time typically exponentially small in $n$.

For an intermediate case, if half the x_i are $(0,1)$ and the other half are $(1,0)$ then typically the modal probability will be $c/n$ (meaning, provided the $p_i$ are bounded away from $0$ and $1$ and are not dependent on $n$ then you will get an answer of this form). To see this, observe that the two coordinates each have modal probability $c/\sqrt{n}$ and they are independent.

I think it should be fairly easy to prove that in this setting the modal probability for any given collection of $x_i$ is minimised when the $p_i$ are all equal (but I did not think that much!). However this is not true in the general setting where the $X_i$ may take many values, for instance if each $X_i$ takes values $(1,0)$, $(0,0)$, $(0,1)$, $(0,2)$,...,$(0,10000)$ then the probability assigned to $(1,0)$ should be much bigger than $1/10000$ to minimise the modal probability.