An elliptic curve $E$ is defined over $\Bbb{Q}$ if only if it is isomorphic over $\overline{\Bbb{Q}}$ to an elliptic curve whose coefficients are in $\Bbb{Q}$.
My question is, what are examples of an elliptic curve which is not defined over $\Bbb{Q}$ ?
Elliptic curve with coefficients not in $\Bbb{Q}$ is not enough, because it might be isomorphic to an elliptic curve with coefficients in $\Bbb{Q}$.
An elliptic curve $E$ over $\overline{\mathbb{Q}}$ can be defined over $\mathbb{Q}$ if and only if its j-invariant $j(E)$ is a rational number. More precisely, $E$ can be defined over a number field $K\subset \overline{\mathbb{Q}}$ if and only if $j(E)\in K$.
As Mathmo123 points out above, you can find many elliptic curves over $\mathbb{Q}(i)$ which can not be defined over $\mathbb{Q}$:
https://www.lmfdb.org/EllipticCurve/2.0.4.1/?field=2.0.4.1&Qcurves=non-Q-curve
But, it is actually quite easy to find examples yourself. Consider the elliptic curve $y^2 = x^3 + Ax + 1$. For which $A$ is the $j$-invariant a rational number?
Additional remark: If $\lambda\in \overline{\mathbb{Q}}\setminus \{0,1]\}$, then the Legendre elliptic curve $E_{\lambda}$ is defind by $y^2 = x(x-1)(x-\lambda)$. If $\lambda$ lies in $K$, then it is obviously defined over $K$.
The $j$-invariant of $E_{\lambda}$ is given by
$$ j(E_{\lambda}) = \frac{256(1-\lambda(1-\lambda))^3}{(\lambda(1-\lambda))^2} = \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}. $$ So, $E_{\lambda}$ is defined over $\mathbb{Q}$ if and only if there is a rational number $q$ such that $$q\lambda^2(1-\lambda)^2 - 256(1-\lambda+\lambda^2)^3 =0. $$ So, choosing $\lambda $ (for example) to be an eleventh root of $2$ obviously gives a non-rational $j$-invariant. (Because this $\lambda$ does not satisfy a polynomial relation over $\mathbb{Q}$ of degree smaller than $11$.) So, this is a relatively easy way of writing down elliptic curves which can not be defined over the rationals.
