Actually you can use this information for predicting whether your solution will pass or not.

Let f(n) denote time your code takes on a given n for a single test case.

Then total time of your code will be f(n_1) + \dots + f(n_T) for all the T test cases.

So constraint "Sum of n over all test cases in one test file does not exceed 10^5" guarantees that n_1 + \dots + n_T is at max 10^5. Using this information, you can predict exactly how much time your code will take.

Assuming f(n) = \mathcal{O}(n):

then You can notice that your code for all the test cases won’t take more than 10^5 operations, because f(n_1) + \dots + f(n_T) = n_1 + \dots + n_T \leq 10^5

Assuming f(n) = \mathcal{O}(n^2):

then You can notice that your code for all the test cases won’t take more than 10^{10} operations, because f(n_1) + \dots + f(n_T) = n_1^2 + \dots + n_T^2. In this case, if there is a single test case with n = 10^5, then your code will take 10^{10} operations in the that test case.

Similarly, you can find desired complexity for any problem in which constraints on sum are given.