Let \$p_i\$ be the probability of the \$i\$-th check succeeding, and let \$N\$ be a random variable describing the number of successful casting attempts. Then the probability of successfully casting the spell at least \$k\$ times is: $$\mathrm P(N ≥ k) = p_1 \times p_2 \times \cdots \times p_k = \prod_{i=1}^k p_i,$$ where the \$\Pi\$ symbol denotes multiplying together \$p_i\$ for values of all \$i\$ from \$1\$ to \$k\$.
Knowing these probabilities for all \$k\$ is enough to fully characterize the distribution of \$N\$. In particular, it lets us calculate the probability of successfully casting the spell exactly \$k\$ times: $$\mathrm P(N = k) = \mathrm P(N ≥ k) - \mathrm P(N ≥ k+1) = (1-p_{k+1}) \prod_{i=1}^k p_i.$$ (This should make sense when you think about it: it's just the probability of succeeding on the first \$k\$ checks and the failing the next one.)
In particular, we can also calculate the expected number of successful casts using either of the following formulas: $$\begin{aligned} \mathbb E(N) &= \sum_{k=1}^\infty k \times \mathrm P(N = k) \\ \mathbb E(N) &= \sum_{k=1}^\infty P(N ≥ k) \end{aligned}$$
These formulas can be fairly easily applied by hand: first calculate a table of the single-roll success probabilities \$p_i\$, keeping in mind the changing die size. Then compile another table of the cumulative probabilities \$\mathrm P(N ≥ k)\$ of succeeding on the first \$k\$ rolls (i.e. the product of the first \$k\$ items in the first table), and finally sum the numbers in the second table to obtain the expectation value \$\mathbb E(N)\$.
Or you can just make a spreadsheet to do it for you.
(Note: the linked spreadsheet is read only. To test it for different spell levels, make a copy first.)
Alternatively, we can calculate the probabilities "backwards" using the following recurrence: $$\begin{aligned} N_{k-1} &= \begin{cases} 1 + N_k & \text{with probability }p_k \\ 0 & \text{with probability }1-p_k \end{cases} \\ &= \mathbf 1_{p_k} (1 + N_k),\end{aligned}$$ where \$N_k\$ is a random variable denoting the number of successful casting attempts after \$k\$ successes, and \$\mathbf 1_{p_k}\$ is a random variable that equals \$1\$ with probability \$p_k\$ and \$0\$ otherwise.
Using this formula, it's quite easy to calculate the distribution of \$N\$ iteratively using AnyDice and plot the results:
DICE: {20, 12, 10, 8, 6, 4:15}
loop LEVEL over {1..9} {
SPELLS: 0
loop X over [reverse DICE] {
SPELLS: (dX > 2*LEVEL) * (1 + SPELLS)
}
output SPELLS named "spell level [LEVEL]"
}
Of course, to apply this "backwards" formula, we have to start with an approximation of \$N_k\$ for some sufficiently large value of \$k\$. The code above effectively assumes that the player will never succeed on more than fifteen d4 rolls (i.e. no more than 20 rolls total), and thus starts with \$N_{20} = 0\$. (You can change this limit by changing the length of the DICE array.)
For spell level 2 and above this assumption is exact (and in fact overkill), since a d4 roll is guaranteed to fail; for level 1 spells it's not, but in practice the probability of succeeding on 15 successive checks with a 50% success rate each is negligible (\$2^{-15} ≈ 0.00003\$), so it's a very good approximation.
