Just run across this question, and am surprised that the first example that came to mind was not mentioned:

Fermat's "Last Theorem" is heuristically true for $n > 3$, but heuristically false for $n=3$ which is one of the easier cases to prove.

if $0 < x \leq y < z \in (M/2,M]$ then $|x^n + y^n - z^n| < M^n$. There are about $cM^3$ candidates $(x,y,z)$ in this range for some $c>0$ (as it happens $c=7/48$), producing values of $\Delta := x^n+y^n-z^n$ spread out on the interval $(-M^n,M^n)$ according to some fixed distribution $w_n(r) dr$ on $(-1,1)$ scaled by a factor $M^n$ (i.e., for any $r_1,r_2$ with $-1 \leq r_1 \leq r_2 \leq 1$ the fraction of $\Delta$ values in $(r_1 M^n, r_2 M^n)$ approaches $\int_{r_1}^{r_2} w_n(r) dr$ as $M \rightarrow \infty$).

This suggests that any given value of $\Delta$, such as $0$, will arise about $c w_n(0) M^{3-n}$ times. Taking $M=2^k=2,4,8,16,\ldots$ and summing over positive integers $k$ yields a rapidly divergent sum for $n<3$, a barely divergent one for $n=3$, and a rapidly convergent sum for $n>3$.

Specifically, we expect the number of solutions of $x^n+y^n=z^n$ with $z \leq M$ to grow as $M^{3-n}$ for $n<3$ (which is true and easy), to grow as $\log M$ for $n=3$ (which is false), and to be finite for $n>3$ (which is true for relatively prime $x,y,z$ and very hard to prove [Faltings]).

More generally, this kind of analysis suggests that for $m \geq 3$ the equation $x_1^n + x_2^n + \cdots + x_{m-1}^n = x_m^n$ should have lots of solutions for $n<m$, infinitely but only logarithmically many for $n=m$, and finitely many for $n>m$. In particular, Euler's conjecture that there are no solutions for $m=n$ is heuristically false for all $m$. So far it is known to be false only for $m=4$ and $m=5$.

Generalization in a different direction suggests that any cubic plane curve $C: P(x,y,z)=0$ should have infinitely many rational points. This is known to be true for some $C$ and false for others; and when true the number of points of height up to $M$ grows as $\log^{r/2} M$ for some integer $r>0$ (the rank of the elliptic curve), which may equal $2$ as the heuristic predicts but doesn't have to. The rank is predicted by the celebrated conjecture of Birch and Swinnerton-Dyer, which in effect refines the heuristic by accounting for the distribution of values of $P(x,y,z)$ not just "at the archimedean place" (how big is it?) but also "at finite places" (is $P$ a multiple of $p^e$?).

The same refinement is available for equations in more variables, such as Euler's generalization of the Fermat equation; but this does not change the conclusion (except for equations such as $x_1^4 + 3 x_2^4 + 9 x_3^4 = 27 x_4^4$, which have no solutions at all for congruence reasons), though in the borderline case $m=n$ the expected power of $\log M$ might rise.

Warning: there are subtler obstructions that may prevent a surface from having rational points even when the heuristic leads us to expect plentiful solutions and there are no congruence conditions that contradict this guess. An example is the Cassels-Guy cubic $5x^3 + 9y^3 + 10z^3 + 12w^3 = 0$, with no nonzero rational solutions $(x,y,z,w)$:

Cassels, J.W.S, and Guy, M.J.T.: On the Hasse principle for cubic surfaces, Mathematika 13 (1966), 111--120.