Equating the zero point energy with the cosmological constant is a common misconception shared even by the most sophisticated physicists.
However, the zero point energy and the cosmological constant are totally different animals.
The energy-momentum tensor $T^{\mu\nu}_\Lambda$ of the cosmological constant is of the form:
$$
T^{00}_\Lambda = \rho_\Lambda,
$$
and
$$
T^{11}_\Lambda = T^{22}_\Lambda = T^{33}_\Lambda= p_\Lambda
$$
with
$$
p_\Lambda = -\rho_\Lambda.
$$
And what does the zero point energy look like? Take a massless fermion for example, the vacuum energy-momentum tensor $T^{\mu\nu}_F$ can be calculated as (see details on page 55 here):
$$
T^{00}_F = \rho_F = -\frac{2\hbar}{(2\pi)^3}\int k d^3k,
$$
and
$$
T^{11}_F = T^{22}_F = T^{33}_F= p_F = -\frac{2\hbar}{3(2\pi)^3}\int k d^3k
$$
Therefore, with a proper regularization/cutoff, one has
$$
p_F = \frac{\rho_F}{3}
$$
which is categorically different from the case of cosmological constant $p_\Lambda = -\rho_\Lambda$.
Hence there is no similarity between the zero point energy and the cosmological constant at all!
In cosmological nomenclature
$$
p = w\rho
$$
where $w$ is called equation of state parameter, which is $-1$ for the cosmological constant and $1/3$ (radiation-like) for the above massless fermion example. Changing to massive fermion/boson will not help the case either (interested reader is encouraged to verify independently).
Added note.
Some may challenge the above calculation. To corroborate the notion let's quote another paper (page 12) Everything You Always Wanted To Know About The Cosmological Constant Problem (But Were Afraid To Ask);
It is clear from the previous expressions that $p/\rho \neq -1$ which indicates that the stress energy tensor is not
of the form ∝ $-\rho g_{\mu\nu}$. In the limit m → 0, as can be
easily shown from Eqs. (75) and (78), the equation of
state is in fact $p/\rho = 1/3$. This would mean that the
zero point fluctuations do not behave like a cosmological
constant but rather like radiation.
The paper goes on to discuss how to fix this with dimensional regularization. But the dimensional regularization usually kills off non-logarithmic divergences and the divergent integral in hand is quartical, so I am not particularly convinced.
More added note.
Let's look at the fermion kinematic Lagrangian (ignoring mass term) in curved space time:
$$
L_{F} \sim \bar{\psi}e\wedge e\wedge e\wedge d\psi
$$
where $e$ is the vierbein/tetrad/frame 1-form and $\wedge$ denote wedge product between differential forms.
And what does the cosmological constant Lagrangian look like? it's
$$
L_{\Lambda} \sim e\wedge e\wedge e\wedge e
$$
As you can see, if the zero point energy of the fermion can be somehow equated to the cosmological constant, you have to magically convert $e\wedge e\wedge e\wedge d$ to $e\wedge e\wedge e\wedge e$, which is mission impossible.
BTW, note that the Higgs potential is of the form:
$$
V_{H} \sim (-m_H^2 |\phi|^2 + \lambda |\phi|^4) e\wedge e\wedge e\wedge e
$$
which may indeed contribute to cosmological constant, should the Higgs field $\phi$ develop a non-zero VEV upon spontaneous symmetry breaking. Mind you that the Higgs potential contribution is a separate story from the zero point energy ppl usually talk about.
