Sleafar got the right answer - 10 - by forcing it. I got same by the following observation: in order to get numbers above 6 one have to make a group of (1+1+1) to get 3 to further operate with. as the remaining we have two "1" and 3 operations available. Among these 3 operations one should be reserved for connection to the first group. And to get number over 7 one must use + operator to get 2. That leads us to solve the problem having 3 and 2 and two remaining operators. Having these assumptions, 8 and 9 are possible to produce while 10 is not.
BTW : the max positive integer is ((1+1+1)↑(1+1))! = 9! = 362880