If you consider $f(u,v)=u^v$, and $u(x)=v(x)=x$, then $x^x=f(x)=f(u(x),v(x))$ an using the multivariate chain rule:
$$\frac{df}{dx}=\frac{\partial f}{\partial u}\frac{du}{dx}+\frac{\partial f}{\partial v}\frac{dv}{dx}=f_u+f_v$$
Here we use the fact that $\frac{du}{dx}=\frac{dv}{dx}=1$. And then again:
$$\frac{d^2f}{dx^2}=\frac{df}{dx}\left(f_u+f_v\right)=f_{uu}+f_{uv}+f_{vu}+f_{vv}=f_{uu}+2f_{uv}+f_{vv}$$
Inductively,
$$\frac{d^nf}{dx^n}=\sum_{k=0}^n\binom{n}{k}f_{\overbrace{u\cdots u}^k\overbrace{v\cdots v}^{n-k}}$$
Now
$$\begin{align}
f_{\overbrace{u\cdots u}^k\overbrace{v\cdots v}^{n-k}}
&=\frac{\partial^{k}}{\partial u^k}u^v(\ln u)^{n-k}\\
&=\sum_{j=0}^k\binom{k}{j}\left[\frac{\partial^{j}}{\partial u^j}u^v\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]\\
&=\sum_{j=0}^k\binom{k}{j}\left[j!\binom{v}{j}u^{v-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]\\
\end{align}$$
It's gets messy after this (but I claim interesting!) so let's recap a bit before continuing. At this point we have:
$$f^{(n)}(x)=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}\left[j!\binom{v}{j}u^{v-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
And we may let $u=v=x$ wherever we do not yet need to take a derivative:
$$f^{(n)}(x)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{x}{j}x^{x-j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
In fact we may let $x=1$ where we do not yet need to take a derivative, since ultimately you are after $f^{(n)}(1)$.
$$f^{(n)}(x)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\left[\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\right]$$
Now we work on $\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}$. We have a composition of functions $g\circ h$, where $h(u)=\ln(u)$ and $g(u)=u^{n-k}$. It's $(k-j)$th derivative can be found with Faà di Bruno's formula.
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!i!^{m_i}}g^{\left(\sum m_i\right)}\left(\ln(u)\right)\prod_t\left(\frac{d^t}{du^t}\ln(t)\right)^{m_t}
\end{align}$$
where the first sum is over tuples of nonnegative integers such that $m_1+2m_2+\cdots+(k-j)m_{k-j}=k-j$. The function $g$ is a power function with nonnegative power $n-k$, so high derivatives of it are zero. Any tuple of the $m_i$ with $\sum m_i>n-k$ will contribute zero to the sum So we can use an additional constraint that $\sum m_i\leq n-k$.
We can simplify more, knowing that $g$ is a power function and knowing derivatives of $\ln(u)$.
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!i!^{m_i}}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}\prod_t\left(-(-1)^t(t-1)!u^{-t}\right)^{m_t}
\end{align}$$
The $i!$ in the denominator have some cancellation with $(t-1)!$ at the end. And then the ultimate power of $u$ is $-1m_1-2m_2-\cdots=-(k-j)$. And the product of all those $(-1)$s can be simplified:
$$\begin{align}
&\frac{\partial^{k-j}}{\partial u^{k-j}}(\ln u)^{n-k}\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}\left[\prod_t\left(-(-1)^t\right)^{m_t}\right]\left[\prod_t\left(u^{-t}\right)^{m_t}\right]\left[\prod_t\frac{\left((t-1)!\right)^{m_t}}{t!^{m_t}}\right]\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(\ln u)^{n-k-\sum m_i}(-1)^{k-j+\sum m_i}u^{j-k}\left[\prod_t\frac{1}{t^{m_t}}\right]\\
&=\sum_{m_1,\ldots,m_{k-j}}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(-1)^{k-j+\sum m_i}u^{j-k}(\ln u)^{n-k-\sum m_i}
\end{align}$$
Regrouping from earlier, and letting $u=x=1$:
$$f^{(n)}(1)=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\sum_{m_i}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(\sum m_i\right)!\binom{n-k}{\sum m_i}(-1)^{k-j+\sum m_i}(0)^{n-k-\sum m_i}$$
Note that last power of $0$ is not $0$ when the exponent is $0$. So we proceed under the added constraint that $\sum m_i=n-k$.
$$
\begin{align}
f^{(n)}(1)&=\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}\left[j!\binom{1}{j}\right]\sum_{m_i}\frac{(k-j)!}{\prod_i i^{m_i}m_i!}\left(n-k\right)!(-1)^{n-j}\\
&=n!\sum_{k=0}^n\sum_{j=0}^k\binom{1}{j}\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}(-1)^{n-j}
\end{align}$$
For $j>1$, the summand is $0$. So we consider the $j=0$ and $j=1$ terms only. But there is no $j=1$ term for $k=0$, so we also separate the $(k,j)=(0,0)$ term:
$$
\begin{align}
f^{(n)}(1)
&=(-1)^nn!\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\\
&+(-1)^nn!\sum_{k=1}^n\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\\
&+(-1)^{n-1}n!\sum_{k=1}^n\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}
\end{align}$$
Note we are only considering $n\geq 1$. We have accumulated some constraints on the tuples of $m_i$:
- It's a $(k-j)$-tuple: $m_1,m_2,\ldots,m_{k-j}$
- $\sum im_i=k-j$
- $\sum m_i=n-k$
In the first of the three terms above, $k=j=0$, so $k-j=0$. So we have an empty sum, $0$.
In the second term above, $j=0$. So we are summing over a $k$-tuple with $\sum im_i=k$.
In the third term, $j=1$. So we are summing over a $(k-1)$-tuple with $\sum im_i=k-1$.
$$
\begin{align}
f^{(n)}(1)=(-1)^nn!\sum_{k=1}^n\left[\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}-\sum_{m_i}\frac{1}{\prod_i i^{m_i}m_i!}\right]
\end{align}$$
In addition to the constraints above on the tuples, we also have $\sum m_i=n-k$.
Let's do check with $n=3$. When $k=1$, the first sum is over a $1$-tuple summing to $2$. So it's just $[2]$ but that does not satisfy $\sum im_i=k$. The second sum is over a $0$-tuple, so empty, contributing $0$.
When $k=2$, the first sum is over a $2$-tuple summing to $1$. The possibilities are $[1;0]$ and $[0;1]$. Only the latter satisfies $\sum im_i=k=2.$ And the second sum is over a $1$-tuple summing to $1$. So $[1]$. This does satisfy $\sum im_1=k-1=1$.
When $k=3$, the first sum is over a $3$-tuple summing to $0$. The only possibility is $[0;0;0]$. That does not satisfy $\sum im_i=k=3.$ And the second sum is over a $2$-tuple summing to $0$. So $[0;0]$. This does not satisfy $\sum im_1=k-1=2$.
So we have:
$$
\begin{align}
f^{(3)}(1)
&=-6\left[\frac{1}{1^{0}0!2^11!}-\frac{1}{1^{1}1!}\right]=3
\end{align}$$
This agrees with what you have for $f^{(3)}(1)$.
So we have a relatively concise expression for $f^{(n)}(1)$. However it sums over $k$, and then depends on partitions $n-k$ that satisfy an additional constraint. If you can simplify it further, that would be neat.
Here is an expression for the Taylor series centered at $1$.
$$T_1(x)=1+\sum_{n=1}^{\infty}(-1)^n\sum_{k=1}^n\left[\sum_{\begin{array}{c}m_1+\cdots+m_k=n-k\\1m_1+\cdots+km_k=k\end{array}}\frac{1}{\prod_{i=1}^k i^{m_i}m_i!}-\sum_{\begin{array}{c}m_1+\cdots+m_{k-1}=n-k\\1m_1+\cdots+(k-1)m_{k-1}=k-1\end{array}}\frac{1}{\prod_{i=1}^{k-1} i^{m_i}m_i!}\right](x-1)^n$$
And if this converges when $x=0$, you have
$$T_1(0)=1+\sum_{n=1}^{\infty}\sum_{k=1}^n\left[\sum_{\begin{array}{c}m_1+\cdots+m_k=n-k\\1m_1+\cdots+km_k=k\end{array}}\frac{1}{\prod_{i=1}^k i^{m_i}m_i!}-\sum_{\begin{array}{c}m_1+\cdots+m_{k-1}=n-k\\1m_1+\cdots+(k-1)m_{k-1}=k-1\end{array}}\frac{1}{\prod_{i=1}^{k-1} i^{m_i}m_i!}\right]$$
But I wouldn't know how to show that the sum converges, or independently find that it converges to $1$.
