does a number that contains all primes less than it exist?

This is impossible. For every $n>1$, we have gcd$(n,n-1)=1$. Since there is a prime dividing $n-1$, the result follows.

Edit : I should have assumed $n>2$ since I need a prime divisor of $n-1$.

To make Lord Shark's comment explicit:

Let $n$ be composite and let $p$ be its largest prime factor. Then certainly $n\ge 2p$. By Bertrand's postulate, there exists a prime $q$ between $p$ and $2p$, hence there exist primes below $n$ that do not divide $n$.

A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n\#$.

Your question in effect asks, "Is it ever the case that $p_{n+1}>p_n\#$?"

We can start by considering the number $p_n\#+1$. Since every one of the first $n$ primes divides $p_n\#$, and none of them divide $1$, we conclude none of them can divide $p_n\#+1$. So that number must either be a prime or have prime factors. If it has prime factors, they must be other than any of the first $p_n$, and those factors must be smaller than $p_n\#$, and larger than $p_n$ itself. In this case, the prime next larger than $p_n$ would have to be smaller than $p_n\#$ and the answer to your question would be "No". So that forces us to consider that $p_n\#+1$ is in fact a prime, and postulate that it is the prime next larger than $p_n$, i.e. $p_{n+1}$. If it were ever the case that $p_{n+1}=p_n\#+1$, that would be the only possible way to answer your question in the positive.

Here is how Levent's answer sinks that ship. We must now consider the number $p_n\#-1$. As in the case of $p_n\#+1$, each $p_n$ divides $p_n\#$ but none of them divide $1$, so $p_n\#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n\#$. That being the case, $p_n\#+1$ cannot be the prime next larger than $p_n$.

Levent's answer stands on its own in answering your question in the negative. However, the excursion taken here shows that Levent's answer also undercuts the only reasonable candidate for the kind of number you postulate.