This is not exactly an answer. I try to give a meaning to $f(x)$. Not as a limit. I will give a formal solution as a power series (possibly not converging).

Consider a function of two parameters $$u(t,x)=(t+1)x^{(t+2)x^{(t+3)x^{(t+4)x^{(t+5)x^{(t+6)x^{(t+7)x^{.{^{.^{.}}}}}}}}}}$$ so that $f(x)=u(0,x)$. Now whatever the meaning we assign to $f(x)$ it appear that $u(t,x)$ satisfies the functional equation $$u(t,x)=(t+1) x^{u(t+1,x)}.$$ So that $$\log u(t,x)=\log(t+1)+u(t+1,x)\log x.$$ It is natural to put $u(t,1)=t+1$. So we try an analytic solution $$u(t,x)=(t+1)\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)$$ Then $$\log\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)= (t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\log x$$ Expanding $$a_1(t)(x-1)+\Bigl(a_2(t)-\frac{1}{2}a_1(t)^2\Bigr)(x-1)^2+ \Bigl(\frac13a_1(t)^3-a_1(t)a_2(t)+a_3(t))(x-1)^3+\cdots =$$ $$= (t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\Bigl((x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots\Bigr)$$ Giving equations $$a_1(t)=t+2$$ $$a_2(t)-\frac{1}{2}a_1(t)^2=-\frac{t+2}{2}+(t+2)a_1(t+1)$$ So $$a_2(t)=\frac12(t+2)^2-\frac{t+2}{2}+(t+2)(t+3)=(t+2)\Bigl(\frac{t+2}{2}-\frac12+t+3 \Bigr)=(t+2)\frac{3t+7}{2}. $$ It is easy to show that the equation resulting from equating the coefficients of $(x-1)^n$ determine $a_n(t)$ in a unique way, as a polynomial of degree $n$.

With Mathematica we obtain $$\eqalign{ a_1(t)&=\frac{1}{1!}(2+t)\cr a_2(t)&=\frac{1}{2!}(14+13t+3t^2)\cr a_3(t)&=\frac{1}{3!}(234+287t+117t^2+16t^3)\cr a_4(t)&=\frac{1}{4!}(6792+10014 t+5505 t^2+1348 t^3+125 t^4)\cr a_5(t)&=\frac{1}{5!}(301980+507614 t+338960 t^2+113200 t^3+19030 t^4+1296 t^5)\cr a_6(t)&=\frac{1}{6!}(18996384+35272036 t+27073062 t^2+11070440 t^3+2559125 t^4+318834 t^5+16807 t^6)\cr }$$ Therefore the formal solution is the power series $$f(x)=1+\frac{2}{1!}(x-1)+\frac{14}{2!}(x-1)^2+\frac{234}{3!}(x-1)^3 +\frac{6792}{4!}(x-1)^4+\frac{301980}{5!}(x-1)^5+\frac{18996384}{6!}(x-1)^6+\cdots $$ The sequence of coefficients do not appear in the OEIS. The coefficient of $t^n$ in $a_n(t)$ appear to be $n^{n-2}$.

I have computed the coefficients of the power series for $f(x)$ for $n\le30$. It appear that $\log a_n(t)\le \frac34 n^{3/2}$, (with approximate equality) if this trend continues, the series will be divergent except for $x=1$. But Euler summed the divergent series $\sum n! x^n$.

This is not exactly an answer. I try to give a meaning to $f(x)$. Not as a limit. I will give a formal solution as a power series (possibly not converging).

Consider a function of two parameters $$u(t,x)=(t+1)x^{(t+2)x^{(t+3)x^{(t+4)x^{(t+5)x^{(t+6)x^{(t+7)x^{.{^{.^{.}}}}}}}}}}$$ so that $f(x)=u(0,x)$. Now whatever the meaning we assign to $f(x)$ it appear that $u(t,x)$ satisfies the functional equation $$u(t,x)=(t+1) x^{u(t+1,x)}.$$ So that $$\log u(t,x)=\log(t+1)+u(t+1,x)\log x.$$ It is natural to put $u(t,1)=t+1$. So we try an analytic solution $$u(t,x)=(t+1)\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)$$ Then $$\log\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)= (t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\log x$$ Expanding $$a_1(t)(x-1)+\Bigl(a_2(t)-\frac{1}{2}a_1(t)^2\Bigr)(x-1)^2+ \Bigl(\frac13a_1(t)^3-a_1(t)a_2(t)+a_3(t))(x-1)^3+\cdots =$$ $$= (t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\Bigl((x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots\Bigr)$$ Giving equations $$a_1(t)=t+2$$ $$a_2(t)-\frac{1}{2}a_1(t)^2=-\frac{t+2}{2}+(t+2)a_1(t+1)$$ So $$a_2(t)=\frac12(t+2)^2-\frac{t+2}{2}+(t+2)(t+3)=(t+2)\Bigl(\frac{t+2}{2}-\frac12+t+3 \Bigr)=(t+2)\frac{3t+7}{2}. $$ It is easy to show that the equation resulting from equating the coefficients of $(x-1)^n$ determine $a_n(t)$ in a unique way, as a polynomial of degree $n$.

With Mathematica we obtain $$\eqalign{ a_1(t)&=\frac{1}{1!}(2+t)\cr a_2(t)&=\frac{1}{2!}(14+13t+3t^2)\cr a_3(t)&=\frac{1}{3!}(234+287t+117t^2+16t^3)\cr a_4(t)&=\frac{1}{4!}(6792+10014 t+5505 t^2+1348 t^3+125 t^4)\cr a_5(t)&=\frac{1}{5!}(301980+507614 t+338960 t^2+113200 t^3+19030 t^4+1296 t^5)\cr a_6(t)&=\frac{1}{6!}(18996384+35272036 t+27073062 t^2+11070440 t^3+2559125 t^4+318834 t^5+16807 t^6)\cr }$$ Therefore the formal solution is the power series $$f(x)=1+\frac{2}{1!}(x-1)+\frac{14}{2!}(x-1)^2+\frac{234}{3!}(x-1)^3 +\frac{6792}{4!}(x-1)^4+\frac{301980}{5!}(x-1)^5+\frac{18996384}{6!}(x-1)^6+\cdots $$ The sequence of coefficients do not appear in the OEIS. The coefficient of $t^n$ in $a_n(t)$ appear to be $(n+1)^{n-1}$.

I have computed the coefficients of the power series for $f(x)$ for $n\le30$. It appear that $\log a_n(t)\le \frac34 n^{3/2}$, (with approximate equality) if this trend continues, the series will be divergent except for $x=1$. But Euler summed the divergent series $\sum n! x^n$.