Yes, even with no assumptions on $f$. Suppose we have an exact sequence
$$0\to M_n\to\cdots\to M_1\to M_0\to0$$
of flat $A$-modules. If $N$ is any $A$-module, then the sequence
$$0\to M_n\otimes_AN\to\cdots\to M_1\otimes_AN\to M_0\otimes_AN\to0$$
is exact. In the case of $f:\mathop{\mathrm{Spec}}B\to\mathop{\mathrm{Spec}}A$, pullback is given by $-\otimes_AB$, so this applies, and the general case reduces to the affine case.
To prove the claim, break the long exact sequence up into short exact sequences. Starting on the right we have
$$0\to K_1\to M_1\to M_0\to0$$
where $K_1=\ker(M_1\to M_0)=\mathrm{coker}(M_3\to M_2)$. Then we get a long exact sequence
$$\mathrm{Tor}_1^A(M_0,N)\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$
But $M_0$ is flat, so the Tor vanishes, giving
$$0\to K_1\otimes_AN\to M_1\otimes_AN\to M_0\otimes_AN\to0.$$
Because $M_1$ is flat as well, the long exact sequence also shows that $\mathrm{Tor}_1^A(K_1,N)=0$, meaning $K_1$ is also flat. Now look at the next short exact sequence
$$0\to K_2\to M_2\to K_1\to0$$
where $K_2=\ker(M_2\to M_1)=\mathrm{coker}(M_4\to M_3)$. We have seen that $K_1$ is flat, so the same argument as above shows that
$$0\to K_2\otimes_AN\to M_2\otimes_AN\to K_1\otimes_AN\to0$$
is exact. And so on.
I gave you a bad hint in the comments because I had not thought about this in a while. Let me try to make up with it by giving you a (hopefully enlightning) solution. The key issue in your attempted solution is that you're not really engaging with the method suggested by Hartshorne: we can define a filtration of $\bigwedge^r\mathcal{F}$ which has successive subquotients $\bigwedge^p \mathcal{F}'\otimes\bigwedge^{r-p}\mathcal{F}''$. We define this filtration locally, and then check that it's "independent enough" to patch together globally.
We're going to use the same strategy as part (c): we're going to define a filtration of $\bigwedge^r \mathcal{F}$,
$$\bigwedge^r\mathcal{F} = F^0\supseteq F^1\supseteq\cdots\supseteq F^r \supseteq F^{r+1}=0 $$
which has successive quotients $F^p/F^{p+1}=\bigwedge^p\mathcal{F}' \otimes \bigwedge^{r-p} \mathcal{F}''$. Considering $r=\operatorname{rank} \mathcal{F}$, this will imply the result, as the filtration will reduce to $\bigwedge^r \mathcal{F} \cong \bigwedge^{rk \mathcal{F}'}\mathcal{F}'\otimes \bigwedge^{rk \mathcal{F}''}\mathcal{F}''$.
The way we construct this filtration is that we set it up in a basis-independent way on open subsets $U\subset X$ where each of $\mathcal{F}',\mathcal{F},\mathcal{F}''$ are free, and then check that this means it glues in to a filtration of the global sheaves. Fix an arbitrary such $U$, and choose any splitting $\mathcal{F}|_U\cong \mathcal{F}'|_U\oplus \mathcal{F}|_U''$. This gives us $$\bigwedge^r\mathcal{F}|_U \cong \bigoplus_{i=0}^r \left(\bigwedge^i \mathcal{F}'|_U\right)\otimes \left(\bigwedge^{r-i} \mathcal{F}''|_U\right).$$
Now we construct the filtration by induction - the idea of our strategy is to set $F^i$ to be those wedge products which have at least $i$ entries coming from a basis of $\mathcal{F}'|_U$. To do this, set $F^{r+1}=0$, and assume we've picked $F^{j+1},\cdots,F^{r+1}$ satisfying the requested properties. To construct $F^j$, consider the image of
$$\varphi: \bigwedge^j\mathcal{F}'|_U \otimes \bigwedge^{r-j} \mathcal{F}''|_U \to \left(\bigwedge^r \mathcal{F}|_U\right)/F^{j+1}.$$
I claim the preimage of this under the projection $\pi:\bigwedge^r\mathcal{F}|_U\to \left(\bigwedge^r\mathcal{F}|_U\right)/F^{j+1}$ is independent of the chosen splitting: pick a basis $x_1,\cdots,x_s$ for $\mathcal{F}'|_U$ and a basis $y_1,\cdots,y_t$ for $\mathcal{F}''|_U$, where we identify each with their image inside $\mathcal{F}|_U$. The image of $\varphi$ is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge y_{q_1}\wedge\cdots\wedge y_{q_{r-j}}$$ as the $p$ range over $1,\cdots,s$ and the $q$ range over $1,\cdots,t$. Then for any other basis $y_1+c_1,\cdots,y_t+c_t$ with $c_i\in \mathcal{F}'|_U$, the image is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge (y_{q_1}+c_{q_1})\wedge\cdots\wedge (y_{q_{r-j}}+c_{q_{r-j}}).$$ But expanding the wedge product in this sum, we see that it differs from a vector in our original collection by wedge products which have at least $j+1$ entries which are selected from the $x$s, and thus this difference is in $F^{j+1}$, so our claim about basis-independence is proven, and we can set $F^j$ to be this preimage.
This means our filtration $F^j$ is basis-independent, and in particular, it's compatible with restriction maps between open subsets where $\mathcal{F},\mathcal{F}',\mathcal{F}''$ are all free since any basis of the sheaves on the larger set is still a basis on the smaller set. Since any scheme is covered by such opens, this means we can patch our filtration together to an honest filtration of sheaves with the specified quotients globally, and we're done.
Best Answer
Computing the determinant of $\Omega^1{X \times T/T}$ with the help of the last two sequences, one concludes that $p_1^*(\mathcal{J}) \cong \mathcal{O}_{X \times T}$, which implies $\mathcal{J} \cong \mathcal{O}_X$.