The game is trivially equivalent to take the 52 cards from the deck, one (or two) at a time, and group the same valued cards in respective 13 piles. You loose if at some moment you have 8 or more piles with an odd number of cards.
My simulation ($5 \times 10^7$ tries) gave me a winning probability estimate of $p=0.054207$
and the amount of cards taken from the deck (minimum=8, maximum=52=win) has the following distribution. Most of the time, we loose in the first trials, as noted in the other answer.
Update: a computer calcutation (not a simulation; exact, and quick, though not elegant ) gives me: $$
p=\frac{23655404966027488102313337962261}{436763610067824320055367170703125}=0.0541606590401477 $$
I just consider a 5-elements "state" (number of values that have appeared k-times) and compute iteratively the probabilities transitions (1806 valid states). Java runnable code, in case you are interested (with floating point instead of exact fractions, for clarity), here.
Edited: Let $t \in 0 \dots 52$ be the "time" (iteration) of the game, equal (in my formulation above) to the total number of drawn cards.
Let ${\bf n}(t) $ be a 5-dimensional state vector such that $n_k(t)\in 0 \dots 13$ (with $k \in 0 \dots 4$) counts how many values (card ranks) have appeared $k$ times, after having drawn $t$ cards. It follows that
$$\sum_{k=0}^4 n_k(t)=13$$
$$\sum_{k=0}^4 k \, n_k(t)=t$$
(BTW, the total set of states, correspond to the weak-compositions of 13 in 5 parts. And the state value determines univocally the time $t$, through the second equation).
The legal state transitions correspond to selecting a index value $j\in 0\cdots 3$, decrementing that index value and incrementing the next one: $n_j(t+1)=n_j(t)-1$, $n_{j+1}(t+1)=n_{j+1}(t)+1$, with the restriction $n_j(t) >0$.
Each transition probability can be computed as
$$p_j = \frac{n_j(t) \, (4-j)}{52-t}$$
Further, the additional game rule corresponds to having less than $M=8$ odd valued numbers, i.e. $n_1(t) + n_3(t) < 8$
Then, given the probability of all legal states at time $t$, we can compute the probability of all legal states at time $t+1$. And the probability of reaching ("legally", without losing the game) state $t$ is the sum of those probabilities. This is just what my program computes.
An example. This states corresponds to 28 cards already played, 3 values (12 cards) not having yet appeared, 5 values have appeared twice (10 cards), 2 values have appeared three times).
k n nk p
----------------
0 3 0 12/24
1 0 0 0/24
2 5 10 10/24
3 2 6 2/24
4 3 12 0/24
----------------
13 28 1