For $a\in[1,e]$ we have that $\ln(a)\in [0,1]$ thus $x_n(a)=\exp^{(n)}(a)$ is a subsequence going to infinity such that $T(x_n(a))$ can be any number in $[0,1]$ depending of the initial value of $a$.
Now you asked this for integers, but $y_n(a)=\lfloor x_n(a)\rfloor$ is an integer subsequence such that $x_n(a)\le y_n(a)<x_n(a)+1$
And since $\ln$ and its iterates are striclty increasing functions we have ($n\gg 1$)
$T(x_n(a))\le T(y_n(a))\lt T(x_n(a)+1)$
edit: this is true if it is the same $i$ in $\log^*$ but the following shows it is.
But $\ln(x_n(a)+1)=\ln(x_n(a))+\ln(1+1/x_n(a))=\ln(x_n(a))+x_n(a)^{-1}+o(x_n(a)^{-1})$
And when we iterate $\ln(\ln(x_na(a)+1)=\ln(\ln(x_n(a))+(x_n(a)\ln(x_n(a))^{-1}+o((x_n(a)\ln(x_n(a))^{-1})$
But according to the process $\ln(x_n(a))\ge 1$ so we can majorate roughly
$\ln(\ln(x_na(a)+1)\le\ln(\ln(x_n(a))+O(x_n(a)^{-1})$ and this goes on as we iterate again.
All that to say that $T(x_n(a)+1)\sim T(x_n(a))$.
Thus $\lim\limits_{n\to\infty} T(y_n(a))=\lim\limits_{n\to\infty} T(x_n(a))=\ln(a)$.
We have an integer subsequence $(y_n(a))_n$ going to infinity such that $T(y_n(a))$ can take any pre-decided value in $[0,1]$ so $\lim\limits_{n\to\infty} T(n)$ doesn't exists.
Edit: I realize that I have worked with a different definition of $log^*$.
In my mind it was $\log^*(x)=\max\{i\in\mathbb N\mid \ln^{(i)}(x)\ge 1\}$, the demo can be adapted to your definition I think.
By your definition $T(n)<0$ while $f(n)\ge 0$ so $f$ can only be an upper bound for $T$.
Consider the following sequences, where $k$ goes to infinity.
$$f(\pi/2 + k2\pi) = \pi/2 + k2\pi \rightarrow +\infty$$
$$f(-\pi/2 + k2\pi) = \pi/2 - k2\pi \rightarrow -\infty$$
You can compute the integral explicitely with an integration by part.
$$ \int_0^T x \sin(x) dx = [-x \cos(x)]_0^T - \int_0^T \cos(x) dx =
-T \cos(T) +\sin(T).$$
Here again you can find sequences for which this last quantity goes to $-\infty$ or $+\infty$.
Best Answer
Let $l_i\in {\mathbb N}$ be nondecreasing. Since $\sum 2^{-l_i}$ is bounded, one has $\lim_{i\rightarrow \infty} l_i=\infty.$ One needs to prove that $$\lim_{n\rightarrow \infty}\frac n{2^{l_n}}=0.\qquad (1)$$ One can prove a more general result:
Proposition. Let $a_n>0$ be a nondecreasing sequence such that $\sum_{n=1}^\infty \frac 1 {a_n}$ converges. Then $$\lim_{n\rightarrow \infty}\frac n{a_n}=0.$$
Proof. For every $\epsilon>0,$ there exists $N_1\in {\mathbb N}$ such that $\sum_{n>N_1}\frac 1{a_n}<\frac{\epsilon}2.$ Choose $N\geq N_1$ such that $\frac {N_1}{a_N}<\frac {\epsilon}2,$ hence $\frac {N_1}{a_n}<\frac {\epsilon}2$ for $n\geq N.$
Then if $n>N,$ one has $$\frac n{a_n}=\frac {N_1}{a_n}+\frac{n-N_1}{a_n}<\frac{\epsilon}2+\sum_{i=N_1+1}^n\frac 1 {a_i}$$ $$<\frac{\epsilon}2+\sum_{i>N_1}\frac 1 {a_i}<\frac {\epsilon}2+\frac{\epsilon}2=\epsilon.$$ This proves the proposition.
Now (1) follows by letting $a_n=2^{l_n}.$