The well-known four fours puzzle asks us to generate as many consecutive positive integers starting with $1$ as we can. In the variation discussed in this question, the only allowed operations are $+,-, \cdot,$ and $\div.$ I have managed to generate the first nine positive integers using all four fours, but we can do better with other sets of numbers. Here's an example: $1, 2, 3, 12:$
$1 = (12 \div (1 + 3)) - 2.$
$2 = (12 \div 2) - 3 - 1.$
$3 = (12 \div 2) - (3 \div 1).$
$4 = (12 \div 2) - 3 + 1.$
$5 = (12 \div 3) - 1 + 2.$
$6 = (12 \div 3) + (2 \div 1).$
$7 = (12 \div 3) + 1 + 2.$
$8 = ((12 \cdot 2) \div 3) \cdot 1.$
$9 = ((12 \cdot 2) \div 3) + 1.$
$10 = 12 - 3 + 2 - 1$
$11 = 12 - 3 + (2 \div 1)$
$12 = 12 - 3 + 2 + 1$
$13 = 12 + 3 - (2 \div 1)$
$14 = 12 + 3 - 2 + 1$
$15 = (12 \div (2 - 1)) + 3$
$16 = 12 + 3 + 2 - 1$
$17 = 12 + 3 + (2 \div 1)$
$18 = 12 + 3 + 2 + 1$
$19 = ((12 \cdot 3) \div 2) + 1$
$20 = (12 \cdot 2) - 3 - 1$
$21 = (12 \cdot 2) - (3 \div 1)$
$22 = (12 \cdot 2) - 3 + 1$
$23 = ((12 + 1) \cdot 2) - 3$
$24 = 12 \cdot (3 - 2 + 1)$
$25 = ((12 - 1) \cdot 2) + 3$
$26 = (12 \cdot 2) + 3 - 1$
$27 = (12 \cdot 2) + (3 \div 1)$
$28 = (12 \cdot 2) + 3 + 1$
$29 = ((12 - 2) \cdot 3) - 1$
$30 = 12 \cdot (3 - (1 \div 2))$
$31 = ((12 - 2) \cdot 3) + 1$
$32 = (12 + 3 + 1) \cdot 2$
$33 = (12 \cdot 3) - 2 - 1$
$34 = (12 \cdot 3) - (2 \div 1)$
$35 = (12 \cdot 3) - 2 + 1$
$36 = 12 \cdot 3 \cdot (2 - 1)$
$37 = (12 \cdot 3) + 2 - 1$
$38 = (12 \cdot 3) + (2 \div 1)$
$39 = (12 \cdot 3) + 2 + 1$
$40 = (12 - 2) \cdot (3 + 1)$
$41 = ((12 + 1) \cdot 3) + 2$
$42 = (12 + 2) \cdot 3 \cdot 1$
$43 = ((12 + 2) \cdot 3) + 1$
As you can see, this set of numbers goes up to $43$. How high can such a set of four numbers go (if we must use all four in each expansion)?
Here's a (very crude) upper bound:
There are $5$ ways to place parentheses, since every expression can be "parenthesized". There are $4$ options for the first operation, the second operation, and the third operation, so the operations have $64$ possibilities. There are also $24$ ways to order the operands. Therefore, $5 \cdot 64 \cdot 24 = 7680$ is an upper bound.
Now that I have time, let me put my comment above into action. The traditional notation we use for binary operators such as $+,-,\times, \div$ is called "infix" notation. That is, the operator is placed between its two arguments. When multiple operations are used, this leads to an ambiguity of meaning that we resolve with parentheses. But in the 19th century, it was recognized that the ambiguity could also be resolved by giving the operator first, then the two arguments. For example, consider $$8=((12\cdot 2)\div 3)\cdot 1$$ in your list. In prefix notation, this becomes $$8 = \times\div\times(12)(2)(3)(1)$$ Where the parentheses are now only needed because to distinguish the atomic values (and only because we use juxtaposition of digits to denote larger numbers). To interpret this, note that each operator takes two arguments. The rightmost $\times$ takes the two values that follow it as its arguments: $$\times(12)(2) = 12 \times 2$$ The division similarly takes two arguments. Its first argument is the rightmost times, so $$\div\times(12)(2)(3) = (\times(12)(2)) \div 3 = (12 \times 2) \div 3 $$
Finally the leftmost $\times$ has $\div\times(12)(2)(3)$ and $(1)$ as its arguments: $$\times\div\times(12)(2)(3)(1) = (\div\times(12)(2)(3)) \times 1 = ((12 \times 2) \div 3)\times 1$$
All expressions in these binary operators can similarly be translated into a prefix form, where parentheses are only used in denoting the 4 values, not the order of operations. Thus there is no need to consider them when counting possible values of the expressions. In forming a prefix expression with three binary operators "$O$" and four atomic values "$A$", there are only a few options are well-formed: $$OOOAAAA\\OOAOAAA\\OOAAOAA\\OAOAOAA$$Anything else would result in an operator being satisfied before all the elements are involved, or in an operator not having two arguments to act on.
So we have $4$ choices of distribution of operators and atoms, $4!$ ways of distributing our four atomic values over the four atom locations, and $4^3$ ways of picking the three operators. But we can reduce the number of distinct cases.
As Hagen points out, with addition and multiplication, it doesn't matter which order the arguments come in. Exchanging them leads to an identical value. Thus for each multiplication or addition used, we can throw out half of the possible cases as duplicates. I argue that we can similarly throw out half the cases for division and subtraction as well. As we are only interested in positive integer values, division or subtraction can only be used in such a way that the final result is positive and integer. If a use of one of these operators does produce a non-integer or negative result, some later operation must fix this. But if this occurs, the expression can always be rearranged so that either the division is replaced with a multiplication/subtraction by an addition, or when the division/subtraction occurs, the result is already a positive integer. This allows us the freedom to throw out the original arrangement, and always choose the second. And for remaining divisions and subtractions, only one of the two arrangements of their arguments will result in a positive integer.
Thus for all operators, we can throw out one of the two arrangements of its arguments as either duplicate or producing an unacceptable value. For three operations, this means only one in eight orderings can produce a unique acceptable result. This gives an upper bound of $$4 \times 4! \times 4^3 \div 2^3 = 768$$ possible values.
Obviously, my argument involves a lot of hand-waving that needs to be thoroughly examined before it can be accepted, but I am fairly confident it will hold true. Additionally, it still counts cases as distinct that will often not be, and as having a valid result when that result will not be valid. So it is still a pretty high upper bound.