There is no built-in function to do this (the Lua standard library is very bare, and ROBLOX doesn't really add anything that deals with plain types like tables)

Here's an outline of how you might do it:

Given your list, build a map count where count[value] is the number of times that value appeared in list.

Implementation

-- returns a `counts` map such that `counts[value]`
-- is the number of times that `value` appears in `list`
function countMap(list)
	local counts = {}
	for _, value in ipairs(list) do
		if counts[value] then	
			-- 2nd or on time that we've seen this element
			counts[value] = counts[value] + 1
		else
			-- 1st time we've seen this element
			counts[value] = 1
		end
	end
	return counts
end

This can be shortened a bit, if you want:

-- returns a `counts` map such that `counts[value]`
-- is the number of times that `value` appears in `list`
function countMap(list)
	local counts = {}
	for _, value in ipairs(list) do
		-- "default" counts[value] to 0 if it's not already there
		counts[value] = (counts[value] or 0) + 1
	end
	return counts
end

After we have the counts, we just need to find the key that has the largest value.

-- returns a `key` such that `map[key]` is as large as possible
function keyLargest(map)
	local best = nil
	for key in pairs(map) do
		if best == nil then
			-- first key in the map
			best = key
		elseif map[best] < map[key] then
			-- `key` has a bigger value than `best`,
			-- so `key` is our new best
			best = key
		end
	end
	return best
end

This can be shortened a bit, if you want:

-- returns a `key` such that `map[key]` is as large as possible
function keyLargest(map)
	local best = next(map) -- get an arbitrary key
	for key in pairs(map) do
		if map[best] < map[key] then
			best = key
		end
	end
	return best
end

Then mostCommon can just be defined in terms of the above two functions:

-- returns an element that appears the most number of times in `list`
function mostCommon(list)
	return keyLargest(countMap(list))
end

Note that this will break ties arbitrarily, i.e., mostCommon{1, 1, 2, 2} could be either 1 or 2.