*
This is my reasoning: If k is the number of ventures and n is the number of coppers (lets keep it simple with just two types of treasure), then in the case where k<<n (if the deck is mostly ventures then the treasure per turn is bounded by the size of the deck, which is linearly related to ventures. If this is what you're talking about, then I agree. If you're in the case where you're playing almost your whole deck, then only Bank or Venture/HoP/Mandarin will yield quadratic treasure).
Then the chance of any given card in hand being a venture is k/(n+k), which is approximately k/n.
Now in the large case, we can sample with replacement. Therefore when a venture is played the ventures hit before copper is geometric, so we expect the value of a venture played from your hand to be 1+ 1/(n/(n+k-1)), which comes out to be 1+(k-1+n)/n
so the value of your hand = expected $ of ventures *expected value of a venture=5*chance of a card being a venture*expected value of a venture. The highest term is k^2 and it is clearly not canceled out anywhere. Therefore it is quadratic.
Now you might ask, well, if I have a venture as one card in my hand, then doesn't that mean less chance of another card being a venture? Yes, but it balances out because if your first card is a venture, you can just not look at your other cards in hand, and play it. Then the venture has a certain expected value, which was calculated. If you have any other ventures in hand, you can forget about the first one, because the first card (be it copper or venture) isn't likely to take out more ventures or coppers from the deck than the pre-existing ratio, and it will thus have the same expectation.
*Ok, I understand what you mean now. The main point is basically that (k being coppers and n being ventures in your formula) n/n+k-1 is approximately n/k since we're taking k to be significantly larger than n. It's up to debate whether we can have an arbitrarily large constant k that is always larger but doesn't grow with n, but what you said before about the value dependence of ventures is not a reason against it being quadratic.
Spoilered for obsolescence
Edit: Ok, I think you're right. The calculation I'm using for this is that given a large deck of v ventures and c coppers, the total treasure of the deck is v+c and it takes c/5 turns to cycle through it, so ventures indeed increase the treasure per turn linearly.