I believe the only way you can do this is to assume you have fixed length inputs to the hash function $f$. Otherwise, it is problematic what probability distribution you'd want to impose on the input set $\{0,1\}^{\ast}$ which is the collection of all finite input strings. In practice, hash functions do have an upper limit on the input string, but that's astronomical, in terms of testing all input strings.
So, let's assume the hash function has security parameter of $k$ bits. This corresponds to the function acting like a random function with outputs of length $n=2k$ bits.
Your testing would generate a large number of random values from a uniform distribution on $\{0,1\}^{m},$ thus treating the hash function as a random function $f:\{0,1\}^m\rightarrow \{0,1\}^n.$ Let this random set of inputs be denoted by $X$. Now define
$$
a_{ij}=\#\{x \in X:[f(x\oplus e_i)]_j \neq [f(x)]_j\}
$$
for $1\leq i\leq n,1\leq j\leq m,$
where $e_i$ is the vector with a one in the $i^{th}$ position and zeroes everywhere and $[u]_j$ denotes the $j^{th}$ component of vector $u$. $a_{ij}$ counts the number of inputs from $X$ which differ in the $j^{th}$ output bit when the $i^{th}$ input bit is flipped.
You can now define a degree of strict avalanche criterion $D_{SAC}(f)$ as
$$
D_{SAC}(f):=1-\frac{\sum_{i=1}^n \sum_{j=1}^n \left|\frac{2a_{ij}}{\#X}-1\right|}{nm},
$$
with the expectation that $D_{SAC}(f)$ should be approximately 1, i.e., the sum of the absolute differences $$\left|\frac{2a_{ij}}{\#X}-1\right|$$ over $i$ and $j$ should be small. One way of expressing this may be to say $E[A_{ij}] \approx (\#X/2)$ if $A_{ij}$ is a random variable representing the $a_{ij}$ with $A_{ij}$ distributed as the binomial variable $Bin(\#X,1/2).$ From here you can then model the overall experiment in terms of chi squared variables, by using the Gaussian approximation to the binomial. So take the properly scaled $nm$ variables as being from a chisquared distribution with $nm-1$ degrees of freedom.