Consider the generating functions $$ R_k(u) = \sum_{j_1,\ldots,j_k\geq 0} u^{j_1 + 2 j_2 + \cdots + k j_k} \prod_{t=1}^k \frac 1{j_t! t^{j_t}} f_t(j_1,\ldots,j_t). $$ I will prove by induction on $k$ that $$ R_k(u) = \frac 1 u + (1-\frac 1u) \exp(u+\frac {u^2}2+\cdots+\frac {u^k}k) = \frac {u^k}{k+1} + O(u^{k+1}). $$
Note that the first formula implies the second one. Indeed, we have $$ (1-\frac 1u) \exp(u+\frac {u^2}2+\cdots+\frac {u^k}k+\frac {u^{k+1}}{k+1}+\cdots) = (1-\frac 1u) \exp(-\ln(1-u)) = -\frac 1u. $$ Terms with $u^{k+2}$ and higher do not contribute to $u^k$ and lower, so we get $$ (1-\frac 1u) \exp(u+\frac {u^2}2+\cdots+\frac {u^k}k+\frac {u^{k+1}}{k+1}) = -\frac 1u +O(u^{k+1}) $$ and $$ (1-\frac 1u) \exp(u+\frac {u^2}2+\cdots+\frac {u^k}k) = -\frac 1u \exp(-\frac {u^{k+1}}{k+1}) +O(u^{k+1}) = -\frac 1u +\frac {u^k}{k+1}+O(u^{k+1}). $$
The base of induction $k=1$ is rather easy, we get $$R_1 = \sum_{j\geq 0}\frac {u^j}{j!} \frac j{j+1}$$ $$=\exp(u) - \sum_{j\geq 0}\frac {u^j}{(j+1)!} = \exp(u) + \frac {\exp(u)-1}u$$ and the statement follows.
To prove the induction step $k\to (k+1)$ observe that $$ R_{k+1}(u) = R_k(u) \exp(\frac {u^{k+1}}{k+1}) - u^k( {\rm Coeff.~of~}R_k{\rm~at~}u^k) \Big(\sum_{j\geq 0} \frac {u^{j(k+1)}}{(j+1)! (k+1)^j}\Big) $$ (using the induction assumption) $$= R_k(u) \exp(\frac{u^{k+1}}{k+1}) -\frac {u^k}{k+1} \Big(\frac{\exp(\frac {u^{k+1}}{k+1})-1}{\frac {u^{k+1}}{k+1}}\Big) $$ $$ =(R_k(u)-\frac 1u)\exp(\frac {u^{k+1}}{k+1}) + \frac 1u $$ $$ =\frac 1 u + (1-\frac 1u) \exp(u+\frac {u^2}2+\cdots+\frac {u^{k+1}}{k+1}). $$
It remains to observe that the coefficient of $R_n(u)$ at $u^n$ is exactly the original formula.