Polynomials can also have complex coefficients. There is actually quite a bit of work related to the topic. See, for example, here or here.
An example of such a polynomial would be
$$(42+42i)x^2+(7+i)x.$$
The roots of the above polynomial are at $$x=0 $$
and
$$ x = -\frac{2}{21} + \frac{i}{14}.$$
So, nobody hinders you to create polynomials with complex coefficients.
I'll prove your form is inner product over $\mathbb C[x]=V$ (polynomial space with base field $\mathbb C$) by proving three properties: left linearity, Hermitian symmetry, positive definiteness.
First, I'll use notation such that $p(x)=\sum_0^np_ix^i\ (p\in V)$, simillar as $q,r\in V$. And consider $a\in\mathbb C$.
Part 1: Left Linearity
$$
\langle p+q,r\rangle=(p+q)(0)\overline{r(0)}+\int_0^1(p+q)'(x)\overline{r'(x)}dx
\\
=p(0)\overline{r(0)}+q(0)\overline{r(0)}+\int_0^1p'(x)\overline{r'(x)}dx+\int_0^1q'(x)\overline{r'(x)}dx
\\
=\left(p(0)\overline{r(0)}+\int_0^1p'(x)\overline{r'(x)}dx\right)+\left(q(0)\overline{r(0)}+\int_0^1q'(x)\overline{r'(x)}dx\right)
=
\langle p,r \rangle+\langle q,r \rangle
$$
$$
\langle ap,q \rangle=(ap)(0)\overline{q(0)}+\int_0^1(ap)'(x)\overline{q'(x)}dx=ap(0)\overline{q(0)}+a\int_0^1p'(x)\overline{q'(x)}dx
\\
=a\left(p(0)\overline{q(0)}+\int_0^1p'(x)\overline{q'(x)}dx\right)=a\langle p,q \rangle
$$
So, $\langle \cdot,\cdot \rangle$ satisfies left linearity.
Part 2: Hermitian Symmetry
Since $\overline{\alpha\overline\beta}=\overline\alpha\beta$...
$$
\langle q,p \rangle=q(0)\overline{p(0)}+\int_0^1q'(x)\overline{p'(x)}dx=\overline{p(0)\overline{q(0)}}+\int_0^1\overline{p'(x)\overline{q'(x)}}dx
\\
=\overline{p(0)\overline{q(0)}+\int_0^1p'(x)\overline{q'(x)}dx}=\overline{\langle p,q \rangle}
$$
So, $\langle \cdot,\cdot \rangle$ satisfies Hermitian Symmetry.
Part 3: Positive definiteness
If $p\ne0$, then $p_0$ and $p'(x)$ cannot be both 0, because $p(x)=p_0+\int_0^xp'(t)dt$. So, $\|p_0\|$ and $\|p'(x)\|$ cannot be both 0. So...
$$
\langle p,p \rangle=p(0)\overline{p(0)}+\int_0^1p'(x)\overline{p'(x)}dx=\|p_0\|^2+\int_0^1\|p'(x)\|^2dx\ne0
$$
So, $\langle \cdot,\cdot \rangle$ is positive definite.
Your $\langle \cdot,\cdot \rangle$ is, so, inner product over $V$.
Best Answer
It is simply the polynomial you get by replacing each $a_i$ by its complex conjugate.
You should think of this as the map induced on $\mathbb{C}[x]$ by the map $\mathbb{C}\to\mathbb{C}$ given by complex conjugation.
Added: How do we know that $\int_0^1 p(x)\overline{p(x)}\,dx$ is positive?
Write each $a_j = \alpha_j + i\beta_j$, with $\alpha_j,\beta_j\in\mathbb{R}$. Then note that $v = q(x)+ir(x)$, where \begin{align*} q(x) &= \alpha_0 + \alpha_1 x + \cdots + \alpha_nx^n\\ r(x) &= \beta_0 + \beta_1x + \cdots + \beta_nx^n. \end{align*} So you have: \begin{align*} \langle v,v\rangle &= \int_0^1 (a_0+a_1x + \cdots + a_nx^n)(\overline{a_0}+\overline{a_1}x + \cdots \overline{a_n}x^n)\,dx\\ &= \int_0^1(q(x)+ir(x))(q(x)-ir(x))\,dx\\ &= \int_0^1\Bigl( \left(q(x)\right)^2 - i^2\left(r(x)\right)^2\Bigr)\,dx\\ &= \int_0^1\Bigl( \left(q(x)\right)^2 + \left(r(x)\right)^2\Bigr)\,dx \end{align*} and since this is the integral of two nonnegative real valued functions, it is nonnegative (and equal to $0$ if and only if both $q(x)$ and $r(x)$ are zero, i.e., if and only if $v=\mathbf{0}$).