Let $K$ be any field. I will give a simple $K$-algebra with IBN and a noninner derivation. My example will be a contracted monoid algebra. These have IBN by @PaceNielsen's nice answer here.
Let $X$ be an infinite set and let $M$ be the monoid with zero given by the presentation $$\langle X\cup X^*\mid x^*y=\delta_{x,y}, x,y\in X\rangle$$ where $X^*$ is a bijective copy of $X$. Let $R=K_0M$ be the contracted monoid algebra. So it has $K$-basis $M\setminus \{0\}$ and the product extends that of $M$, where we identify the zero of $M$ with the zero of $K$. This is the Leavitt path algebra of an infinite bouquet of circles and is well known to be simple. This was first proved by Douglas Munn (W. D. Munn. Simple contracted semigroup algebras. In Proceedings of the Conference on Semigroups
in Honor of Alfred H. Clifford (Tulane Univ., New Orleans, La., 1978), pages 35–43. Tulane Univ.,
New Orleans, La., 1979) to the best of my knowledge.
I claim that it has a noninner derivation. This paper studies outer derivations of Leavitt path algebras and the derivation I am using is from there, although technically that paper does not allow digraphs with infinite out-degree.
It is easy to check that every nonzero element of $M$ can be written uniquely in the form $pq^*$ where $p,q$ are words in $X$ (possibly empty) and you extend $\ast$ to words in the obvious way, so that $*$ is an involution on $M$.
Now define $d\colon R\to R$ on the basis $M\setminus \{0\}$ as follows. Fix $x\in X$ and put $$d_x(pq^*) = (|p|_x-|q|_x)pq^*.$$ Here $|w|_x$ is the number of occurrences of the letter $x$ in $w$.
It is easy to check that $d_x$ is a derivation. It is also not too hard to check that it is not inner using that $d_x(x)=x$ and you cannot find $m\in R$ with $xm-mx=x$.
Details added (update).
I found a little time to add in some details. First of all the presentation of $M$ is a complete rewriting system (it is length reducing and there are no overlaps of rules) and the nonzero reduced elements are the elements $pq^*$ with $p,q$ words in $X$.
To see that $R=K_0M$ is simple, let $I$ be a nonzero ideal and let $$0\neq a=\sum_{m\in M}c_mm\in I.$$
We shall use the following observation. Use $|w|$ for the length of a word $w$ over $X$. If $p,q,w,a,b$ are words in $X$ and $0\neq pq^*w=sb^*w$ with $|b|,|q|\geq |w|$, then $pq^*=sb^*$. Indeed, we must have $q=wz$ and $b=wy$ for the product to be nonzero and then $pq^*w = pz^*$ and $sb^*w = sy^*$, whence $p=s$ and $z=y$, i.e., $q=b$.
First suppose that $supp(a)$ contains $1$. Since $X$ is infinite, we can find a letter $x\in X$ such that $x$ appears in no $p,q$ with $pq^*\in supp(a)$. Then $x^*ax=x^*1x=1$ as $x^*pq^*x=0$ whenever $x$ does not appear in $p, q$ and at least one of $p$ or $q $ is nonempty. Next we show that we can modify $a$ to contain $1$ in its support.
Suppose that some word $p$ over $X$ is in the support of $a$ and assume $p$ has minimal length with this property. Then $p^*a$ has $1$ in its support and it is nonzero since $p^*wz^*=1$ in $M$ implies $z=1$ and $w=p$, and so the coefficient of $1$ in $p^*a$ is the coefficient of $p$ in $a$. Thus we are in the previous case, and hence done.
So suppose that all elements of the support of $a$ are of the from $pq^*$ with $q$ nonempty. Choose $pq^*$ in the support of $a$ with $|q|$ minimal. By our observation above, the coefficient of $p$ in $aq$ is the same as the coefficient of $pq^*$ in $a$ because $|q|\leq |b|$ for any $sb^*$ in the support of $a$ and so $sb^*q\neq pq^*q=p$ by the observation. This puts us in the previous case, and hence we are done.
It now follows that $R$ is simple.
Next lets check that $d$ is a derivation. Call a mapping $\theta\colon M\setminus \{0\}\to G$ with $G$ a group a partial homomorphism if $\theta(mn)=\theta(m)\theta(n)$ whenever $mn\neq 0$. For example, there is a partial homomorphism $\theta\colon M\to F_X$ (where $F_X$ is the free group on $X$) with $\theta(pq^*) = pq^{-1}$, and this is in fact the universal partial homomorphism from $M$ to a group, although we don't need this. Now there is a homomorphism $\eta\colon F_X\to \mathbb Z$ sending our fixed $x$ to $1$ and $X\setminus \{x\}$ to $0$. Then $\nu:=\eta\theta\colon M\to \mathbb Z$ is a partial homomorphism with $\nu(pq^*) = |p|_x-|q|_x$.
Therefore, $d(m) = \nu(m)m$ for $m\in M$, where we consider $\nu(0)0=0$ (even though technically $\nu(0)$ is not defined). Then we easily check that, for $m,n\in M$, we have
$$d(mn) = \nu(mn)mn = (\nu(m)+\nu(n))mn = m\nu(n)n+\nu(m)mn = md(n)+d(m)n$$ where this works out even if $mn=0$.
Finally, we observe that $d$ is not inner. Indeed, if $d$ is inner then $x=d(x)=xr-rx$ for some $r\in R$. Note that since $1$ commutes with $x$, we may assume without loss of generality that $1\notin supp(r)$. Since left multiplication by $x$ is injective on $M$ (because $x^*x=1$), we cannot have $x\in x\cdot supp(r)$ because $1\notin supp(r)$. Thus $x\in supp(r)x$. But the only way that can happen is if $xx^*\in supp(r)$ since if $pq^*x=x$ and $p\neq 1$, then $q=p=x$ (use the observation). Let $n\geq 1$ be maximal with $(x^n)(x^n)^*\in supp(r)$. Then $xr$ has $(x^{n+1})(x^n)^*$ in its support (since left multiplication by $x$ is injective) and hence $(x^{n+1})(x^n)^*\in supp(r)x$ since $xr-rx=x$. But if $pq^*\in supp(r)$ with $pq^*x = (x^{n+1})(x^n)^*$ with $n\geq 1$, then $|q|\geq 2$ and the only possibility is $p=x^{n+1}$, $q=x^{n+1}$ by the observation. Thus $x^{n+1}(x^{n+1})^*\in supp(r)$, contradicting the choice of $n$. We deduce that $d$ is not inner.
