I've now got a counterexample to the full question.

Let $B$ be the Boolean algebra consisting of the bounded and bounded-complement regular open subsets of $\mathbb{R}$, where for $\neg a$ we take the interior of the complement of $a$.

Let $A$ be the subalgebra of such sets $a$, such that whenever an integer $n\in a$, then also $(n-1,n+1)\subset a$.

Note that $A$ is quasi-dense in $B$, since if $b$ is bounded, then it is contained in some large interval $(-k,k)$, which is in $A$. And if $b$ has bounded complement, then $b$ contains some $(k,k+1)$ for some large $k$, and this is in $A$.

Let $C$ be the algebra of all regular open subsets of $\mathbb{R}$. Note that $B$ is dense in $C$, since every nonempty regular open set contains an interval.

So we have atomless Boolean algebras $A\subset B\subset C$, and $A$ is quasi-dense in $B$, which is dense in $C$.

But I claim that $A$ is not quasi-dense in $C$. To see this, let $z=\bigcup_{k\in\mathbb{Z}}(k-\frac12,k+\frac12)$. Note that $z$ contain no nonempty element of $A$, since the intervals are too narrow. Similarly, $z$ is not contained in any non-trivial element of $A$, since the only element of $A$ containing every integer is the whole of $\mathbb{R}$.

I was led to the counterexample above by first considering the following instance, which would be a counterexample, except that it is atomic.

Well, this still doesn't answer the atomless question, but I've got a violation of the desired implication among atomic Boolean algebras.

The answer is no. For a counterexample, let $A$ have at least two atoms; let $B$ split one of those atoms, and let $C$ split both of them.

I've now got a counterexample to the full question.

Let $B$ be the Boolean algebra consisting of the bounded and bounded-complement regular open subsets of $\mathbb{R}$, where for $\neg a$ we take the interior of the complement of $a$.

Let $A$ be the subalgebra of such sets $a$, such that whenever an integer $n\in a$, then also $(n-1,n+1)\subset a$.

Note that $A$ is quasi-dense in $B$, since if $b$ is bounded, then it is contained in some large interval $(-k,k)$, which is in $A$. And if $b$ has bounded complement, then $b$ contains some $(k,k+1)$ for some large $k$, and this is in $A$.

Let $C$ be the algebra of all regular open subsets of $\mathbb{R}$. Note that $B$ is dense in $C$, since every nonempty regular open set contains an interval.

So we have atomless Boolean algebras $A\subset B\subset C$, and $A$ is quasi-dense in $B$, which is dense in $C$.

But I claim that $A$ is not quasi-dense in $C$. To see this, let $z=\bigcup_{k\in\mathbb{Z}}(k-\frac12,k+\frac12)$. Note that $z$ contain no nonempty element of $A$, since the intervals are too narrow. Similarly, $z$ is not contained in any non-trivial element of $A$, since the only element of $A$ containing every integer is the whole of $\mathbb{R}$.

I was led to the counterexample above by first considering the following instance, which would be a counterexample, except that it is atomic.

Let $A$ consist of the finite or cofinite subsets of $\mathbb{N}$ that take $2k$ and $2k+1$ together, if at all. That is, $a\in A$ if $a\subset\mathbb{N}$ is finite or cofinite and for every $k$ we have $2k\in a\leftrightarrow 2k+1\in a$. Let $B$ be the Boolean algebra consisting of all finite or cofinite subsets of $\mathbb{N}$.

Note that $A$ is quasi-dense in $B$, since if $b$ is finite, then $b$ is contained in an interval $[0,2k+1]$ for some large $k$, and this is in $A$, and if $b$ is cofinite, then $b$ contains some final segment interval $[2k,\infty)$, which is in $A$.

Let $C$ be the algebra generated by the elements of $B$ together with the set $E$ of even numbers. Thus, every element of $C$ is the union of a finite or cofinite subset of $E$ with a finite or cofinite subset of $\mathbb{N}-E$. The algebra $B$ is dense in $C$, since the singletons are dense, and they are finite.

Finally, $A$ is not quasi-dense in $C$, because the set $E$ contains no nonzero element of $A$, as it contains no odd numbers, and is contained in no non-unital element of $A$, as the only element of $A$ containing all the even numbers is the whole of $\mathbb{N}$.

This is my original answer, which shows merely that quasi-density is not transitive.

The answer is no. For a counterexample, let $A$ have at least two atoms; let $B$ split one of those atoms, and let $C$ split both of them.