A random 13-card hand is dealt from a standard deck of cards. What is the probability that the hand contains at least 3 cards of every suit?
Choose one of the 4 suits, within that suit choose 13 available cards (without replacement,where order doesn't matter).
$4 \choose 1$ * $13 \choose 3$
Choose one of the remaining three suits, within that suit, choose 13 available cards.
$3 \choose 1$ * $13 \choose 3$
Same for the remaining two suits
$2 \choose 1$ * $13 \choose 3$
And the last suit available
$1 \choose 1$ * $13 \choose 3$
At this point I've picked $3*4=12$ cards. Hence I have one more left to pick. Which I pick from a pool of unpicked cards $52-12=40$
So the total number of ways in which I could get dealt a 13-card hand where I have at least 3 cards of each suit is:
$4 \choose 1$ * $13 \choose 3$ * $3 \choose 1$ * $13 \choose 3$ * $2 \choose 1$ * $13 \choose 3$ * $1 \choose 1$ * $13 \choose 3$ $ * 40$
Divide that by the total number of ways I can pick 13 cards from 52 ($52 \choose 13$)
$$ \frac{\binom{4}{1} * \binom{13}{3} * \binom{3}{1} * \binom{13}{3} * \binom{2}{1} * \binom{13}{3} * \binom{1}{1} * \binom{13}{3} * 40}{\binom{52}{13}} $$
What is wrong with my approach above?
Your denominator is correct. For the numerator, choose the suit from which four cards are drawn, select four cards from that suit, and select three cards each from each of the other suits, which can be done in $$\binom{4}{1}\binom{13}{4}\binom{13}{3}\binom{13}{3}\binom{13}{3}$$ distinguishable ways.
As David pointed out in the comments, the order in which the suits are selected does not matter, so there is no need to multiply by the $4!$ ways of selecting the suits. However, it does matter from which suit four cards are selected. There are four ways to select this suit. Also, as lulu pointed out in the comments, designating a card as the fourth card from a suit overcounts by a factor of $4$ since any of the four cards in that suit could be your additional card. Notice that selecting $5\spadesuit, 9\spadesuit, J\spadesuit$ as your three spades and $2\spadesuit$ as your additional card results in the same hand as selecting $2\spadesuit, 5\spadesuit, 9\spadesuit$ as your three cards and $J\spadesuit$ as your additional card.