Is it possible to construct an ordered field which is also algebraically closed?

No, we cannot do it.

No field of positive characteristic can be totally ordered: since $1$ is a square, it must be greater than $0$, and then so are $1$, $1+1$, $1+1+1$, etc; but if the characteristic is $p\gt 0$, then $1+1+\cdots+1$ with $p$ summands is both positive and equal to $0$ which is impossible.

So a totally ordered field must be of characteristic zero. In particular, $-1\neq 1$.

Since $1\gt 0$, then $-1\lt 0$. If your field contains a root of $x^2+1$, call it (for lack of a better name) $i$, then $i^2=-1$; but this means that $-1$ is positive, because it is a square. This contradicts the fact that $-1$ is negative. So no totally ordered field can contain a root of $x^2+1$, let alone be algebraically closed.

Suppose by contradiction that you have such an order.

Pick an element $a<0$. The equation $x^2=a$ has a root. But then $x<0$, $x=0$ or $x>0$, all implying that $x^2 \geq 0$.