Why does this convolution of the prime counting function $\pi$ look like a parabola? In this previous question it is shown that the convolution of the prime counting function $\pi$ with itself, is related to the Goldbach conjecture:
$$\pi^*(n):=\sum_{k=0}^n \pi(k) \pi(n-k)$$
The Goldbach conjecture might be written as:
$$\forall n \ge 2: \frac{\pi^*(2n)+\pi^*(2n-2)}{2} > \pi^*(2n-1)$$
I have plotted the function $\pi^*$ and it looks like a parabola:

Question: Is there any explanation for this observation or is this superficial observation maybe wrong?
Here is some SageMath Code to compute parameters $a,b,c,d,e$ for regression:
$$an^4+bn^3+cn^2+dn+e \approx \pi^*(n)$$
Thanks for your help!
 A: It is straightforward to show that
$$\pi^*(n)\sim\frac{n^3}{6\log^2 n}.\tag{$\ast$}\label{ast}$$
So the graph of $\pi^*(n)$ does not look like a parabola. Instead, it looks like the graph of $\frac{n^3}{\log^2 n}$.
Indeed, we have
$$\sum_{\min(k,n-k)<\frac{n}{\log n}} \pi(k) \pi(n-k)\ll\frac{n}{\log n}\pi(n)^2\ll\frac{n^3}{\log^3 n},$$
and also
\begin{align*}\sum_{\min(k,n-k)\geq\frac{n}{\log n}} \pi(k) \pi(n-k)&\sim\frac{1}{\log n^2}\sum_{\min(k,n-k)\geq\frac{n}{\log n}}k(n-k)\\&\sim\frac{1}{\log n^2}\sum_{k=0}^n k(n-k)\\&\sim\frac{n^3}{6\log^2 n}.\end{align*}
Here we used that for $\min(k,n-k)\geq\frac{n}{\log n}$, both $\log k$ and $\log(n-k)$ are asymptotically $\log n$.
Added. The asymptotic formula \eqref{ast} converges rather slowly. Here are some numeric data:
\begin{align*}
\pi^*(10^2)&=16329\\
\pi^*(10^3)&=6311273\\
\pi^*(10^4)&=3119183737\\
\pi^*(10^5)&=1817310193749\\
\pi^*(10^6)&=1181102034701650\\
\pi^*(10^7)&=827525141442938787\\
\pi^*(10^8)&=611768346585852887680
\end{align*}
The corresponding ratios of the two sides of \eqref{ast} are:
\begin{align*}
n=10^2\quad \rightsquigarrow\quad 2.0778\\
n=10^3\quad \rightsquigarrow\quad 1.8069\\
n=10^4\quad \rightsquigarrow\quad 1.5876\\
n=10^5\quad \rightsquigarrow\quad 1.4453\\
n=10^6\quad \rightsquigarrow\quad 1.3526\\
n=10^7\quad \rightsquigarrow\quad 1.2899\\
n=10^8\quad \rightsquigarrow\quad 1.2455
\end{align*}
