I believe it follows from two things:
(1) In a Universe with only a positive cosmological constant $\Lambda>0$, $\rho+3p \propto \Lambda-3\Lambda = -2\Lambda < 0$.
(2) In a Universe with matter+radiation+a cosmological constant, at late times the cosmological constant will dominate the energy density budget of the Universe, according to the Friedman equation.
More details
The stress-energy tensor for a cosmological constant is
\begin{equation}
T^{(\Lambda)}_{\mu\nu} = -\frac{\Lambda}{8\pi G} g_{\mu\nu}
\end{equation}
where $g_{\mu\nu}$ is the metric.
This follows from starting from Einstein's equations with a cosmological constant
\begin{equation}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}
\end{equation}
and moving the $\Lambda$ term to the opposite side:
\begin{equation}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = 8\pi G \left( T_{\mu\nu} - \frac{\Lambda}{8\pi G} g_{\mu\nu} \right) = 8\pi G \left( T_{\mu\nu} + T^{(\Lambda)}_{\mu\nu} \right)
\end{equation}
Under the symmetries of cosmology (homogeneity and isotropy), in FLRW coordinates, the metric (assuming zero spatial curvature) is
\begin{equation}
ds^2 = -dt^2 + a(t)^2 d\vec{x}^2
\end{equation}
where $a(t)$ is the scale factor, and the stress energy tensor of a perfect fluid can be parameterized as $T_{\mu\nu}=(\rho+p)u_\mu u_\nu + p g_{\mu\nu}$, or in terms of components (with $u^\mu$ identified with the time direction in FLRW coordinates)
\begin{equation}
T_{\mu \nu} = \begin{pmatrix}
\rho & & & \\
& a^2 p & & \\
& & a^2 p & \\
& & & a^2 p
\end{pmatrix}
=
-\frac{\Lambda}{8\pi G} \begin{pmatrix}
-1 & & & \\
& a^2 & & \\
& & a^2 & \\
& & & a^2
\end{pmatrix}
\end{equation}
Therefore, we identify $\rho=-p=\frac{\Lambda}{8\pi G}$.
The trace of the stress energy tensor is
\begin{equation}
T = g^{\mu\nu}T_{\mu\nu} = - \rho + 3 p
\end{equation}
The strong-energy condition states that for every timelike vector $X^\mu$, the following condition is satisfied
\begin{equation}
\left(T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu}\right) X^\mu X^\nu \geq 0
\end{equation}
We can simply consider the time direction to define our timelike vector. Then in our context, the strong energy condition becomes
\begin{equation}
T_{00} - \frac{1}{2} T g_{00} = \rho + \frac{1}{2} \left(-\rho + 3 p\right) = \frac{1}{2} \rho + \frac{3}{2} p
\end{equation}
For a cosmological constant with $p=-\rho$,
\begin{equation}
T_{00} - \frac{1}{2} T g_{00} = - \rho = -\frac{\Lambda}{8\pi G} < 0
\end{equation}
which violates the SEC.
