NB: Note that my $a_k$ have different signs from those defined in the question. For me,
$$
(z - z_1)(z-z_2)(z-z_3) = z^3 - a_1\ z^2 + a_2\ z - a_3,
$$
so that $a_k$ is the $k$-th elementary symmetric function of the $z_i$. This doesn't really affect the answer in any significant way.

While I don't think that the final result relating $V$ to the $a_k$ and $\bar a_k$ is that interesting or useful, people might want to know a way to derive it. Here is what I did:
Start with the formula
$$
V=\frac{\sqrt{-1}}{4}\det
\begin{pmatrix}
1&1&1 \\
z_1& z_2& z_3 \\
\overline{z}_1&\overline{z}_2&\overline{z}_3\end{pmatrix},
$$
and note that multiplying the matrix on the right by its transpose yields
$$\begin{pmatrix}
1&1&1 &\\
z_1&z_2&z_3\\
\bar z_1& \bar z_2 &\bar z_3 \end{pmatrix}
\begin{pmatrix}
1&z_1&\bar z_1\\
1&z_2&\bar z_2\\
1&z_3&\bar z_3 \end{pmatrix}
=\begin{pmatrix}
3&a_1&\bar a_1 \\
a_1&{a_1}^2-2a_2& S \\
\bar a_1&S&{\bar a_1}^2-2\bar a_2
\end{pmatrix}
$$
where $S = z_1\bar z_1 + z_2\bar z_2 + z_3\bar z_3$. Thus, one has the polynomial relation
$$
R := 16V^2 - 3S^2 + 2 a_1\bar a_1 S + {a_1}^2{\bar a_1}^2
- 4{a_1}^2\bar a_2 - 4{\bar a_1}^2a_2 + 12 a_2 \bar a_2 = 0.
$$
It remains to find a relation between $S$ and the $a_k$ and $\bar a_k$. To do this, note that $S$ will be a root of the polynomial
$$
Q := \prod_{\pi\in S_3} \bigl(S - z_1{\bar z_{\pi(1)}} - z_2{\bar z_{\pi(2)}} - z_3{\bar z_{\pi(3)}}\bigr).
$$
Note that $Q$ is a polynomial of degree $6$ in $S$ that is symmetric in the $z_i$ and the $\bar z_i$ separately. Hence, $Q$ can be regarded as a polynomial of degree $6$ in $S$ with coefficients that are polynomials in the $a_k$ and $\bar a_k$. The resulting expression for $Q$ as a polynomial in $S$, the $a_k$, and the $\bar a_k$ has $66$ terms. (Actually expressing $Q$ this way is not easy by hand. However, it's very easy to implement the algorithm for writing a symmetric polynomial in three variables as a polynomial in the elementary symmetric functions in those variables on a computer, which is what I did.)

Finally, $R$ and $Q$ are polynomials in $S$ with coefficients that are polynomials in the variables $V, a_1,a_2,a_3,\bar a_1,\bar a_2,\bar a_3$. Set
$$
P(V,a_1,a_2,a_3,\bar a_1,\bar a_2,\bar a_3) := \text{Resultant}_S(R,Q).
$$
Then $P=0$ is the desired relation. Computation (using Maple) shows that it has the following properties: $P$ is irreducible, is even and of degree $12$ in $V$, is of total degree $20$, and contains $598$ monomial terms, with typical integer coefficient in the millions.

Added remark: The formulae simplify dramatically if one assumes that the centroid of the triangle is at $z=0$, i.e., that $a_1 = \bar a_1 = 0$. (This is analogous to the way that the cubic formula itself simplifies when one removes the quadratic term. Moreover, this can always be easily arranged in the usual way by translation.) When $a_1 = \bar a_1 = 0$, we have
$$
R = 16V^2 - 3S^2 + 12 a_2 \bar a_2
$$
and
$$
\begin{array}
\\
Q &= S^6-6{a_2}{\bar a_2}S^4-27{a_3}{\bar a_3}S^3+9{a_2}^2{\bar a_2}^2S^2\\
&\qquad {} +81{a_2}{a_3}{\bar a_2}{\bar a_3}S
-27{a_3}^2{\bar a_2}^3 - 27{a_2}^3{\bar a_3}^2 - 4{a_2}^3{\bar a_2}^3.
\end{array}
$$
The formula for $P$ is still not that nice, though; it has $17$ terms, with coefficients in the millions. Rather than input it here, I'll just recommend that those interested compute the resultant of this reduced $R$ and $Q$ with respect to $S$ to get it.

Another added remark (the real case): If, in addition, one assumes that the $a_k$ are real (and that $a_1=0$), then the polynomial $P$ factors as
$$
16V^2 (16V^2{+}9{a_2}^2)^2
(4096V^6{+}4608{a_2}^2 V^4{+}1296{a_2}^4V^2{-}19683{a_3}^4{-}2916{a_2}^3{a_3}^2).
$$
In particular, note that $V=0$ is always a root and that $P=0$ has a positive real root $V$ if and only if $27{a_3}^2+4{a_2}^3$ is positive (and $a_3\not=0$), i.e., if and only if the original cubic has only one real (nonzero) root, just as one would expect.

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