Bound on the tail of the sum of independent and identically distributed (iid) variables This interesting question was asked at https://math.stackexchange.com/questions/231455/estimator-for-sum-of-independent-and-identically-distributed-iid-variables a while ago but got no answers.  The original author is Jesko Hüttenhain.
Consider the Chernoff bound described in Theorem 1 of this paper:

Theorem 1. Let $X_1,\ldots,X_n$ be discrete, independent random variables such that $E[X_i] = 0$ and $|X_i|<1$ for all $i$. Let $X:=\sum_{i=1}^n X_i$ and $\sigma^2$ be the variance of $X$. Then,
  $$\Pr\left[|X|\ge \lambda\sigma\right] \le 2e^{-\lambda^2/4}$$
  for any $0\le\lambda\le 2\sigma$.

I want to apply this estimator for a computation, but the variance of the variables $X_i$ is unknown to me. Apart from that, my variables satisfy all the conditions of Theorem 1. In fact, my variables are independent and identically distributed (iid).
On the other hand, there is Chebyshev's inequality for finite samples which does not require knowledge of the variance (or mean) of a given random variable and replaces it with the corresponding values of my given sample.
 However, the estimator is not good enough for my application and I was wondering if there is an estimator similar to Theorem 1 but not featuring the variance of the distribution itself. 
Intuitively speaking, it would be nice to have the best of both worlds: If that is not possible, then what is the best bound I can achieve for a sum of iid variables, without any knowledge about their variance?
 A: I would use Hoeffding's inequality.  It exploits the boundedness of the random variables (like your Theorem 1), only without the variance making an appearance, and you still get exponential decay in the probability estimate (unlike Chebyshev's inequality).
For a big-picture perspective on the different types of concentration inequalities available, check out Terry Tao's notes on the subject.
A: Inequality (4.6) in the paper at https://arxiv.org/abs/math/0701806 (a shortened version of which is at https://projecteuclid.org/euclid.aos/1176325373 ) implies, in your terms, that 
\begin{equation}
 P(|X|\ge x\sqrt n)\le\frac{2e^3}9\,P(|Z|\ge x)
 \le B(x):=\frac{4e^3}9\,\min\Big(\frac12,\frac1{x\sqrt{2\pi}}\Big)e^{-x^2/2}
\end{equation}
for $x\ge0$, where $Z\sim N(0,1)$. This bound, $B(x)$, is an improvement  of Hoeffding's bound $2e^{-x^2/2}$ for $x>\frac{2e^3}{9\sqrt{2\pi}}=1.780\dots$. On the other hand, the Chebyshev bound $\frac1{x^2}$ is better (that is, smaller) than $2e^{-x^2/2}$ for $x\in(x_1,x_2)$, where $x_1=0.845\dots$ and $x_2\approx2.07\dots$, so that the bound $\min(\frac1{x^2},B(x))$ is better than Hoeffding's bound for $x>0.85$.
