Here, I tried to find any general formula for the question above. I did find the number of even entries in the nth row using this formula: $$[(n+1) - 2^{B(n)}]$$ where $B(n)$ is the sum of the bits in the binary representation of $n$.
From that formula, can I find the number of entries in the nth row divisible by 4??
This is not an answer but a hint. Let $n=\overline{a_r\dots a_0}$, $k=\overline{b_r\dots b_0}$ and $n-k=\overline{c_r\dots c_0}$. Let $f(n;p)$ be the largest number $k$ such that $p^k|n!$ for a prime $p$. This is given by: $$ f(n,p)=n-\sum_{i=0}^r a_i. $$ Therefore the power of $p$ in $\binom nk$ is given by: $$ g(n,k,p)=\sum_{i=0}^r b_i+\sum_{i=0}^r c_i-\sum_{i=0}^r a_i. $$ You can argue yourself that this power is equal to the number of carries when $k$ and $n-k$ are added and this is Kummer's theorem. The only problem is counting $k$ with two carries in their sum which seems to be not clear to me now.
Luca's theorem can shed a light on the number of even entries in the row $n$. The first conclusion can be that when $n=\overline{1\dots 1}=2^k-1$ for some $k$, all the entries are odd. Or when $n$ is a power of $2$, $n-2$ entries are even.