A closely related topic is that of concentration inequalities, which give you a bound (of the sort that you are looking for) which also depends on the number of samples (among other things).
Concretely, the concept of Rademacher complexity is a standard tool to address this sort of problems. The Rademacher complexity can be understood as a permutation test, where you changes your labels randomly. When employed to the problem of estimating the mean, the bound tells you how likely is that you get close to the actual mean by chance (how concentrated are the samples around the mean, and thus, how stable are your estimates based on different samples).
To be more specific, for a sample, $X=(x_{i})$, of size $l$, drawn i.i.d. from a probability distribution, $D$, and for a real-valued function class, $F$, with domain $X$, the empirical Rademacher complexity is the random variable defined as,
$$
\hat{R}_{l}(F) = E_{\sigma}\left[sup_{f \in F}\left|\frac{2}{l}\sum_{i=1}^{l}\sigma_{i}f(x_{i})\right|X\right]
$$
where $\sigma = (\sigma_{1},...,\sigma_{l})$ are independent uniform $\pm1$-valued random variables. The Rademacher complexity is,
$$
R_{l}(F) = E_{S \sim D}[\hat{R}_{l}(F)] = E_{S\sigma}\left[sup_{f \in F}\left|\frac{2}{l}\sum_{i=1}^{l}\sigma_{i}f(x_{i})\right|X\right]
$$
The sup means that it looks for the highest correlation possible with random noise. Now, this concept is relevant because of the following theorem,
Given the above conditions, assuming that $F$ is the class of mappings from $X$ to the interval $[0,1]$, and let $(z_{i})$ be a sample of size $l$. If you fix $\delta \in (0,1)$, then with probability $1-\delta$ over random draws of size $l$, every $f \in F$ satisfies,
$$
E[f(z)] \leq \hat{E}[f(z)] + R_{l}(F) + \sqrt{\frac{ln(2/\delta)}{2l}}
\leq \hat{E}[f(z)] + \hat{R}_{l}(F) + 3\sqrt{\frac{ln(2/\delta)}{2l}}
$$
Notice that the hat is used to indicate the empirical expectation measured on a particular sample.
The idea is to find such a family of f´s and use the theorem. Since $D$ has a compact support, you know that $(W-E[W])^{2}/R$ is bounded in $[0,1]$, where $R$ is the radius of the ball.
Using the properties of the Rademacher complexity and a second theorem which gives you the Rademacher complexity for linear prediction (details can be found here and in great detail here), you get the following bound for your probability
$$
\sqrt{\frac{2R^{2}}{l}}\left(\sqrt2 + \sqrt{ln\frac{1}{\delta}}\right)
$$
P.S. I just realized you referred to the p-norm. But still, you can use the Khintchine inequality to bound that quantity with the 2-norm.