There are many nice identities playing here:
$$\sum_{j+k=u}\frac{1}{jk}=\frac{2H_{u-1}}{u}\tag{1},$$
$$\sum_{h=1}^{n}\frac{H_h}{h}=\frac{1}{2}\left(H_n^2+H_n^{(2)}\right)\tag{2},$$
$$\sum_{h=1}^{n}\frac{H_{h-1}}{h}=\frac{1}{2}\left(H_n^2-H_n^{(2)}\right)\tag{3},$$
$$\sum_{j=1}^{n}\frac{H_j}{n+1-j}=H_{n+1}^2-H_{n+1}^{(2)}\tag{4}.$$
Proofs:
$$(1)\quad \sum_{j+k=u}\frac{1}{jk}=\sum_{j=1}^{u-1}\frac{1}{j(u-j)}=\frac{1}{u}\sum_{j=1}^{u-1}\left(\frac{1}{j}+\frac{1}{u-j}\right)=\frac{2H_{u-1}}{u}.$$
$$(2,3)\quad H_n^2=H_n^{(2)}+2\sum_{j=1}^{n}\frac{1}{j}\sum_{k<j}\frac{1}{k}=H_n^{(2)}+2\sum_{j=1}^{n}\frac{H_{j-1}}{j}=-H_n^{(2)}+2\sum_{j=1}^{n}\frac{H_{j}}{j}.$$
By defining $H_0=0$, we have:
$$(4)\quad\sum_{j=1}^{n}\frac{H_j}{n+1-j}=[x^{n+1}]\frac{\log^2(1-x)}{1-x},$$
but we know the series coefficients of $\log^2(1-x)$, so:
$$(4)\quad\sum_{j=1}^{n}\frac{H_j}{n+1-j}=\sum_{k=1}^{n+1}\frac{2H_{k-1}}{k}$$
and $(4)$ follows from $(3)$. Now we play a bit with the last identity that Greg Martin proved:
$$I_n = \frac{1}{2}\sum_{i+j+k\leq n+3}\frac{1}{ijk}=\frac{1}{2}\sum_{t=3}^{n+3}\sum_{i+j+k=t}\frac{1}{ijk}=\frac{1}{2}\sum_{h=3}^{n+3}\sum_{i=1}^{t-2}\frac{1}{i}\sum_{j+k=t-i}\frac{1}{jk};$$
using $(1)$ we have:
$$I_n=\sum_{t=3}^{n+3}\sum_{i=1}^{t-2}\frac{H_{t-i-1}}{i(t-i)}=\sum_{t=3}^{n+3}\frac{1}{t}\sum_{i=1}^{t-2}H_{t-i-1}\left(\frac{1}{i}+\frac{1}{t-i}\right),$$
then re-indexing the inner sum:
$$I_n=\sum_{t=3}^{n+3}\frac{1}{t}\sum_{i=1}^{t-2}H_{i}\left(\frac{1}{t-i-1}+\frac{1}{i+1}\right),$$
where:
$$\sum_{i=1}^{t-2}\frac{H_i}{i+1}=\sum_{i=1}^{t-1}\frac{H_{i-1}}{i}=\frac{1}{2}\left(H_{t-1}^2-H_{t-1}^{(2)}\right)$$
by $(3)$ and
$$\sum_{i=1}^{t-2}\frac{H_i}{t-1-i}=H_{t-1}^2-H_{t-1}^{(2)}$$
by $(4)$. By putting all together we get:
$$I_n=\frac{3}{2}\sum_{t=3}^{n+3}\frac{H_{t-1}^2-H_{t-1}^{(2)}}{t}.\tag{5}$$
This gives that $I_n$ behaves like $\frac{1}{2}\log^3 n$.
The part below this line is outperformed by my next answer.
Moreover, summation by parts gives:
$$\sum_{k=1}^{m+1}\frac{H_{k-1}^{(2)}}{k}=H_{m}^{(2)}H_{m+1}-\sum_{j=1}^m \frac{H_j}{j^2},$$
$$\sum_{k=1}^{m+1}\frac{H_{k-1}^{2}}{k}=H_{m}^{2}H_{m+1}-\sum_{j=1}^m \frac{(H_j+H_{j-1})H_j}{j}=H_{m}^{2}H_{m+1}-\sum_{j=1}^m \frac{(2H_{j-1}+1/j)(H_{j-1}+1/j)}{j},$$
so:
$$\sum_{k=1}^{m+1}\frac{H_{k-1}^{2}}{k}=H_{m}^{2}H_{m+1}-2\sum_{j=1}^{m}\frac{H_{j-1}^2}{j}-3\sum_{j=1}^m\frac{H_{j-1}}{j^2}-H_{m}^{(3)},$$
and rearranging we get:
$$\sum_{k=1}^{m+1}\frac{H_{k-1}^{2}}{k}=\frac{1}{3}H_{m}^{2}H_{m+1}+\frac{2H_m^2}{3(m+1)}-\sum_{j=1}^m\frac{H_{j-1}}{j^2}-\frac{1}{3}H_m^{(3)}.$$
By summing everything, we have that $(5)$ can be written in the following form:
$$I_{n}=\frac{1}{2}H_{n+2}^2 H_{n+3}+\frac{H_{n+2}^2}{n+3}-\frac{3}{2}H_{n+2}^{(2)}H_{n+3}+H_{n+2}^{(3)}+\frac{3}{2}\sum_{j=1}^{n+2}\frac{H_j}{j^2}.\tag{6}$$
The last sum is now clearly bounded by an absolute constant ($\zeta(3)$, for istance), and I strongly believe that it does not simplify further (I was wrong).
