We know that $-\sum\limits_{d|n}\mu(d)\log d=\Lambda(n)$. Using this we can obtain $$(\Lambda*\Lambda)(n)=\Lambda(n)\log n+\sum\limits_{d|n}\mu(d)\log^2d.$$ In general if I write Dirichlet convolution $k$ times $$A_k=\Lambda*\Lambda*.....*\Lambda$$, then how do we evaluate it. I have reached the following recursion
$$A_k=-\sum_{d|n}\mu(d)\log d (A_{k-1}*u)(\frac{n}{d}).$$ How do we proceed further in such type of situation.
A couple of things:
First, it looks like your stated formula for $\Lambda(n)$ is incorrect. I believe that $\Lambda \ast 1 = \log \implies \Lambda = \log \ast \mu$ so that $$\Lambda(n) = \sum_{d|n} \log(n/d) \mu(d) = - \sum_{d|n} \log(d) \mu(d).$$ Perhaps this was a typo?
I would suggest you take a more analytic approach to the problem since it looks like a direct, or even combinatorial method gets very difficult, very quickly (see this post for reference). Let $\Lambda_k := \Lambda \ast \cdots \ast \Lambda$ denote the $k$-fold convolution of $\Lambda(n)$ with itself. Then we know that the Dirichlet series of $\Lambda_k$ is given by $$\sum_{n \geq 1} \frac{\Lambda_k(n)}{n^s} = (-1)^k \zeta^{\prime}(s)^k / \zeta(s)^k.$$ Now by applying Perron's formula to the right-hand-side of this Dirichlet series we obtain a formula for the summatory functions $\sum_{n \leq x}^{\prime} \Lambda_k(n)$ for any integer $x$. Take the difference of this formula at $x$ and $x-1$ to obtain your $k$-fold convolution sum.
Update (3/6/2021):
There seems to be a direct way to evaluate the summatory function using an adaptation of the proof of Theorem 7.17 in Section 7.4 of Montgomery and Vaughan. The referenced theorem considers asymptotics (uniformly if $|z| \leq R$) of $D_z(x) := \sum_{n \leq x} d_z(n)$ where $x \geq 2$ and the $d_z(n)$ are coefficients of the DGF expansion of $$\zeta(s)^{z} = \prod_p (1-p^{-s})^{-z} = \sum_{n \geq 1} \frac{d_z(n)}{n^s}, \Re(s) > 1.$$ They obtain that $$D_z(x) = \frac{x (\log x)^{z-1}}{\Gamma(z)} + O\left(x (\log x)^{\Re(z)-2}\right), x \geq 2.$$ I am not going to go into the full detail of replicating the technical proof for this case, but instead note that it's there for reference in the MV textbook if you want to see it.
Let's just sketch the bound for the main term of the summatory function $L_k(x) := \sum_{n \leq x} \ell_k(n)$ for $x \geq 2$ where we write $$\left(-\frac{\zeta^{\prime}(s)}{\zeta(s)}\right)^k = \sum_{n \geq 1} \frac{\ell_k(n)}{n^s}, \Re(s) > 1.$$ My back-of-the-envelope calculation shows that the main term of $$\left(-\frac{\zeta^{\prime}(s)}{\zeta(s)}\right)^k = \left(\frac{(s-1)^{-1} + O|s-1|}{1 + O|s-1|}\right)^k \overset{s = 1 + w/\log x}{\longrightarrow} O\left(\frac{w}{\log x}\right)^{-2k},$$ so that when we evaluate the main term of the Hankel contour integral towards the last 2/3 of the MV proof, we get that $$L_k(x) = \frac{1}{2\pi\imath} \int_{\mathcal{C_2}} \left(-\frac{\zeta^{\prime}(s)}{\zeta(s)}\right)^k \frac{x^s}{s} ds = O\left(x (\log x)^{2k-1} \times \frac{1}{2\pi\imath} \int_{\mathcal{H}_2} w^{-2k} e^w dw\right) = O\left(\frac{x (\log x)^{2k-1}}{(2k-1)!}\right), k \in \mathbb{Z}^{+}.$$ Now we can take differences to bound the $\ell_k(x)$ for large $x \geq 2$ as follows: $$L_k(x)-L_k(x-1) = O\left(\frac{1}{(2k-1)!}\left[x (\log x)^{2k-1} - (x-1) (\log x + o(1))^{2k-1}\right]\right) = O\left(\frac{(\log x)^{2k-1}}{(2k-1)!}\right).$$ It's not precise, but the method certainly provides an upper bound on the growth of the divisor sums in this question.